1
For a body performing simple harmonic motion, which one of the following statements is correct?
A
The maximum kinetic energy is directly proportional to the frequency.
B
The time for one oscillation is directly proportional to the frequency.
C
The speed at any instant is directly proportional to the displacement.
D
The maximum acceleration is directly proportional to the amplitude.
(Total 1 mark)
2
Which one of the following graphs shows how the acceleration, a, of a body moving with simple
harmonic motion varies with its displacement, x?
(Total 1 mark)
3
A particle oscillates with undamped simple harmonic motion. Which one of the following
statements about the acceleration of the oscillating particle is true?
A
It is least when the speed is greatest.
B
It is always in the opposite direction to its velocity.
C
It is proportional to the frequency.
D
It decreases as the potential energy increases.
(Total 1 mark)
4
A mass M hangs in equilibrium on a spring. M is made to oscillate about the equilibrium position
by pulling it down 10 cm and releasing it. The time for M to travel back to the equilibrium position
for the first time is 0.50 s. Which line, A to D, is correct for these oscillations?
Page 1 of 50
amplitude/cm
period/s
A
10
1.0
B
10
2.0
C
20
2.0
D
20
1.0
(Total 1 mark)
5
The diagram below shows how the kinetic energy of a simple pendulum varies with
displacement.
(a)
Sketch on the diagram above a graph to show how the potential energy of the pendulum
varies with displacement.
(2)
(b)
(i)
State the amplitude of the oscillation.
...............................................................................................................
(1)
(ii)
The frequency of vibration of the pendulum is 3.5 Hz. Write down the equation that
models the variation of position with time for the simple harmonic motion of this
pendulum.
...............................................................................................................
(1)
(iii)
Calculate the maximum acceleration of the simple pendulum.
(2)
(Total 6 marks)
Page 2 of 50
6
The diagrams show the variation of velocity and acceleration with time for a body undergoing
simple harmonic motion.
Which one of the following is proportional to the change in momentum of the body during the
time covered by the graphs?
A
The area enclosed by the velocity-time graph and the time axis
B
The gradient of the velocity-time graph at the point P
C
The area enclosed by the acceleration-time graph and the time axis
D
The gradient of the acceleration-time graph at the point Q
(Total 1 mark)
7
A particle is oscillating with simple harmonic motion described by the equation:
s = 5 sin (20πt)
How long does it take the particle to travel from its position of maximum displacement to its mean
position?
A
B
C
D
(Total 1 mark)
Page 3 of 50
8
(a)
Simple harmonic motion may be represented by the equation
a = – (2πf)2x
(i)
Explain the significance of the minus sign in this equation.
...............................................................................................................
...............................................................................................................
(1)
(ii)
In Figure 1 sketch the corresponding v-t graph to show how the phase of velocity v
relates to that of the acceleration a.
(1)
(b)
(i)
A mass of 24 kg is attached to the end of a spring of spring constant 60 N m–1. The
mass is displaced 0.035 m vertically from its equilibrium position and released. Show
that the maximum kinetic energy of the mass is about 40 mJ.
(5)
Page 4 of 50
(ii)
When the mass on the spring is quite heavily damped its amplitude halves by the end
of each complete cycle. On the grid of Figure 2 sketch a graph to show how the
kinetic energy, Ek, of the mass on the spring varies with time over a single period.
Start at time, t = 0, with your maximum kinetic energy. You should include suitable
values on each of your scales.
(3)
(Total 10 marks)
9
A particle of mass m executes simple harmonic motion in a straight line with amplitude A and
frequency f. Which one of the following expressions represents the total energy of the particle?
A
2 π2 mfA2
B
2 π2 mf 2A2
C
4 π2 m2f 2A
D
4 π2 mf 2A2
(Total 1 mark)
10
(a)
State the conditions necessary for a mass to undergo simple harmonic motion.
........................................................................................................................
........................................................................................................................
(2)
Page 5 of 50
(b)
A child on a swing oscillates with simple harmonic motion of period 3.2 s.
acceleration of free fall = 9.8 m s–2
(i)
Calculate the distance between the point of support and the centre of mass of the
system.
(2)
(ii)
The total energy of the oscillations is 40 J when the amplitude of the oscillations is
0.50 m. Sketch a graph showing how the total energy of the child varies with the
amplitude of the oscillations for amplitudes between 0 and 1.00 m. Include a suitable
scale on the total energy axis.
(2)
(Total 6 marks)
11
Which one of the following statements is true when an object performs simple harmonic motion
about a central point O?
Page 6 of 50
A
The acceleration is always away from O.
B
The acceleration and velocity are always in opposite directions.
C
The acceleration and the displacement from O are always in the same direction.
D
The graph of acceleration against displacement is a straight line.
(Total 1 mark)
12
A body moves with simple harmonic motion of amplitude A and frequency
What is the magnitude of the acceleration when the body is at maximum displacement?
A
zero
B
4π2Ab2
C
Ab2
D
(Total 1 mark)
13
A body is in simple harmonic motion of amplitude 0.50 m and period 4π seconds. What is the
speed of the body when the displacement of the body is 0.30 m?
A
0.10ms−1
B
0.15ms−1
C
0.20 m s−1
D
0.40 m s−1
(Total 1 mark)
14
A body is in simple harmonic motion of amplitude 0.50 m and period 4π seconds. What is the
speed of the body when the displacement of the body is 0.30 m?
A
0.10 m s–1
B
0.15 m s–1
C
0.20 m s–1
D
0.40 m s–1
(Total 1 mark)
Page 7 of 50
15
A body moves in simple harmonic motion of amplitude 0.90 m and period 8.9 s. What is the
speed of the body when its displacement is 0.70 m?
A
0.11 m s–1
B
0.22 m s–1
C
0.40 m s–1
D
0.80 m s–1
(Total 1 mark)
Page 8 of 50
16
The top graph is a displacement/time graph for a particle executing simple harmonic motion.
Which one of the other graphs shows correctly how the kinetic energy, Ek, of the particle varies
with time?
(Total 1 mark)
Page 9 of 50
17
A particle, whose equilibrium position is at Q, is set into oscillation by being displaced to P,
50 mm from Q, and then released from rest. Its subsequent motion is simple harmonic, but
subject to damping. On the first swing, the particle comes to rest momentarily at R, 45 mm from
Q.
During this first swing, the greatest value of the acceleration of the particle is when it is at
A
P.
B
Q.
C
R.
D
P and R.
(Total 1 mark)
18
Which one of the following gives the phase difference between the particle velocity and the
particle displacement in simple harmonic motion?
A
rad
B
rad
C
D
rad
2π rad
(Total 1 mark)
Page 10 of 50
19
The frequency of a body moving with simple harmonic motion is doubled. If the amplitude
remains the same, which one of the following is also doubled?
A
the time period
B
the total energy
C
the maximum velocity
D
the maximum acceleration
(Total 1 mark)
20
Which one of the following graphs shows how the acceleration, a, of a body moving with simple
harmonic motion varies with its displacement, x?
(Total 1 mark)
21
Which one of the following gives the phase difference between the particle velocity and the
particle displacement in simple harmonic motion?
A
rad
B
rad
C
D
rad
2π rad
(Total 1 mark)
Page 11 of 50
22
(a)
A body is moving with simple harmonic motion. State two conditions that must be satisfied
concerning the acceleration of the body.
condition 1 ....................................................................................................
......................................................................................................................
condition 2 …..........................……...............................................................
......................................................................................................................
(2)
(b)
A mass is suspended from a vertical spring and the system is allowed to come to rest.
When the mass is now pulled down a distance of 76 mm and released, the time taken for
25 oscillations is 23 s.
Calculate
(i)
the frequency of the oscillations,
.............................................................................................................
.............................................................................................................
(ii)
the maximum acceleration of the mass,
.............................................................................................................
.............................................................................................................
(iii)
the displacement of the mass from its rest position 0.60 s after being released.
State the direction of this displacement.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
Page 12 of 50
(c)
Figure 1
Figure 1 shows qualitatively how the velocity of the mass varies with time over the first
two cycles after release.
(i)
Using the axes in Figure 2, sketch a graph to show qualitatively how the
displacement of the mass varies with time during the same time interval.
Figure 2
(ii)
Using the axes in Figure 3, sketch a graph to show qualitatively how the potential
energy of the mass-spring system varies with time during the same time interval.
Figure 3
(4)
(Total 12 marks)
Page 13 of 50
23
A body moves with simple harmonic motion of amplitude A and frequency
.
What is the magnitude of the acceleration when the body is at maximum displacement?
A
zero
B
4π2Ab2
C
Ab2
D
(Total 1 mark)
24
Which graph, A to D, shows the variation of the kinetic energy, Ek, with displacement x for a
particle performing simple harmonic motion?
(Total 1 mark)
Page 14 of 50
25
A body moves with simple harmonic motion of amplitude A and frequency
What is the magnitude of the acceleration when the body is at maximum displacement?
A
zero
B
4π2Ab2
C
Ab2
D
(Total 1 mark)
26
A body moves with simple harmonic motion of amplitude 0.90 m and period 8.9 s. What is the
speed of the body when its displacement is 0.70 m?
A
0.11 m s–1
B
0.22 m s–1
C
0.40 m s–1
D
0.80 m s–1
(Total 1 mark)
Page 15 of 50
27
An object oscillating in simple harmonic motion has a time period T. The first graph shows how
its displacement varies with time. Which of the subsequent graphs, A to D, show how the kinetic
energy, Ek, of the object varies with time?
(Total 1 mark)
28
A particle of mass 0.20 kg moves with simple harmonic motion of amplitude 2.0 × 10–2 m. If the
total energy of the particle is 4.0 × 10–5J, what is the time period of the motion?
A
B
C
D
(Total 1 mark)
Page 16 of 50
29
The graph shows the variation in displacement with time for an object moving with simple
harmonic motion.
What is the maximum acceleration of the object?
A
0.025 m s–2
B
0.99 m s–2
C
2.5 m s–2
D
9.8 m s–2
(Total 1 mark)
30
Which one of the following statements is true when an object performs simple harmonic motion
about a central point O?
A
The acceleration is always directed away from O.
B
The acceleration and velocity are always in opposite directions.
C
The acceleration and the displacement from O are always in the same direction.
D
The graph of acceleration against displacement is a straight line.
(Total 1 mark)
Page 17 of 50
31
The two diagrams in the figure below show a student before and after she makes a bungee jump
from a high bridge above a river. One end of the bungee cord, which is of unstretched length 25
m, is fixed to the top of a railing on the bridge. The other end of the cord is attached to the waist
of the student, whose mass is 58 kg. After she jumps, the bungee cord goes into tension at point
P. She comes to rest momentarily at point R and then oscillates about point Q, which is a
distance d below P.
BEFORE
(a)
(i)
AFTER
Assuming that the centre of mass of the student has fallen through a vertical distance
of 25 m when she reaches point P, calculate her speed at P.
You may assume that air resistance is negligible.
answer = ........................... ms–1
(2)
Page 18 of 50
(ii)
The bungee cord behaves like a spring of spring constant 54 Nm–1.
Calculate the distance d, from P to Q, assuming the cord obeys Hooke’s law.
answer = ................................ m
(2)
(b)
As the student moves below P, she begins to move with simple harmonic motion for part of
an oscillation.
(i)
If the arrangement can be assumed to act as a mass-spring system, calculate the
time taken for one half of an oscillation.
answer = ................................ s
(2)
(ii)
Use your answers from parts (a) and (b)(i) to show that the amplitude A, which is the
distance from Q to R, is about 25 m.
(3)
(c)
Explain why, when the student rises above point P, her motion is no longer simple
harmonic.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 19 of 50
(d)
(i)
Where is the student when the stress in the bungee cord is a maximum?
...............................................................................................................
...............................................................................................................
(1)
(ii)
The bungee cord has a significant mass. Whereabouts along the bungee cord is the
stress a maximum? Explain your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 14 marks)
32
A particle of mass 5.0 × 10–3 kg, moving with simple harmonic motion of amplitude 0.15 m, takes
47 s to make 50 oscillations.
What is the maximum kinetic energy of the particle?
A
2.0 × 10–3 J
B
2.5 × 10–3 J
C
3.9 × 10–3 J
D
5.0 × 10–3 J
(Total 1 mark)
Page 20 of 50
33
A body executes simple harmonic motion. Which one of the graphs, A to D, best shows the
relationship between the kinetic energy, Ek, of the body and its distance from the centre of
oscillation?
(Total 1 mark)
34
(a)
(i)
Name the two types of potential energy involved when a mass–spring system
performs vertical simple harmonic oscillations.
...............................................................................................................
...............................................................................................................
(1)
Page 21 of 50
(ii)
Describe the energy changes which take place during one complete oscillation of a
vertical mass-spring system, starting when the mass is at its lowest point.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(b)
Figure 1 shows how the total potential energy due to the simple harmonic motion varies
with time when a mass-spring system oscillates vertically.
Figure 1
time / s
(i)
State the time period of the simple harmonic oscillations that produces the
energy–time graph shown in Figure 1, explaining how you arrive at your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 22 of 50
(ii)
Sketch a graph on Figure 2 to show how the acceleration of the mass varies with
time over a period of 1.2 s, starting with the mass at the highest point of its
oscillations. On your graph, upwards acceleration should be shown as positive and
downwards acceleration as negative. Values are not required on the acceleration
axis.
Figure 2
(2)
(c)
(i)
The mass of the object suspended from the spring in part (b) is 0.35 kg. Calculate the
spring constant of the spring used to obtain Figure 1. State an appropriate unit for
your answer.
spring constant ...................................... unit ....................................
(3)
(ii)
The maximum kinetic energy of the oscillating object is 2.0 × 10–2 J. Show that the
amplitude of the oscillations of the object is about 40 mm.
(4)
(Total 14 marks)
Page 23 of 50
35
A particle of mass m oscillates in a straight line with simple harmonic motion of constant
amplitude. The total energy of the particle is E. What is the total energy of another particle of
mass 2m, oscillating with simple harmonic motion of the same amplitude but double the
frequency?
A
E
B
2E
C
4E
D
8E
(Total 1 mark)
36
(a)
Figure 1 shows how the kinetic energy, Ek, of an oscillating mass varies with time when it
moves with simple harmonic motion.
Figure 1
(i)
Determine the frequency of the oscillations of the mass.
frequency of oscillation .............................................. Hz
(2)
(ii)
Sketch, on Figure 1, a graph showing how the potential energy of the mass varies
with time during the first second.
(2)
Page 24 of 50
(b)
Figure 2 shows a ride called a ‘jungle swing’.
Figure 2
The harness in which three riders are strapped is supported by 4 steel cables. An advert for the
ride states that the riders will be released from a height of 45 m above the ground and will then
swing with a period of 14.0 s. It states that they will be 1.0 m above the ground at the lowest point
and that they will travel at speeds of ‘up to 120 km per hour’.
(i)
Treating the ride as a simple pendulum, show that the distance between the pivot and
the centre of mass of the riders is about 49 m.
(2)
Page 25 of 50
(ii)
The riders and their harness have a total mass of 280 kg.
Calculate the tension in each cable at the lowest point of the ride, assuming that the
riders pass through this point at a speed of 120 km h–1. Assume that the cables have
negligible mass and are vertical at this point in the ride.
tension in each cable ................................................ N
(4)
(iii)
Show that the maximum speed stated in the advert is an exaggerated claim.
Assume that the riders are released from rest and neglect any effects of air
resistance.
(4)
(iv)
The riders lose 50% of the energy of the oscillation during each half oscillation. After
one swing, the speed of the riders as they pass the lowest point is 20 m s–1.
Calculate the speed of the riders when they pass the lowest point, travelling in the
same direction after two further complete oscillations.
speed of riders .......................................... ms–1
(3)
(Total 17 marks)
Page 26 of 50
37
(a)
A simple pendulum is given a small displacement from its equilibrium position and performs
simple harmonic motion.
State what is meant by simple harmonic motion.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
(i)
Calculate the frequency of the oscillations of a simple pendulum of length 984 mm.
Give your answer to an appropriate number of significant figures.
frequency ....................................... Hz
(3)
(ii)
Calculate the acceleration of the bob of the simple pendulum when the displacement
from the equilibrium position is 42 mm.
acceleration .................................... ms–2
(2)
Page 27 of 50
(c)
A simple pendulum of time period 1.90 s is set up alongside another pendulum of time
period 2.00 s. The pendulums are displaced in the same direction and released at the
same time.
Calculate the time interval until they next move in phase. Explain how you arrive at your
answer.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
time interval .......................................... s
(3)
(Total 10 marks)
38
The frequency of a body moving with simple harmonic motion is doubled. If the amplitude
remains the same which of the following is also doubled?
A
The time period.
B
The total energy.
C
The maximum velocity.
D
The maximum acceleration.
(Total 1 mark)
39
A particle oscillates with undamped simple harmonic motion.
The acceleration of the particle
A
is always in the opposite direction to its velocity.
B
decreases as the potential energy increases.
C
is proportional to the frequency.
D
is least when the speed is greatest.
(Total 1 mark)
Page 28 of 50
40
Which line, A to D, in the table gives the amplitude and frequency of a body performing simple
harmonic motion whose displacement x at time t is given by the equation x = P cos Qt?
Amplitude
Frequency
B
P
2πQ
C
P
D
2P
A
(Total 1 mark)
Page 29 of 50
Mark schemes
1
2
3
4
5
D
[1]
D
[1]
A
[1]
B
[1]
(a)
Max to zero to max with zero at 0 displacement and correct amplitude correct shape drawn
with reasonable attempt to keep total energy constant, crossing at 1 × 10–2 J
A1
(2)
(b)
(i)
0.044 m
B1
(1)
(ii)
x = 0.044 cos 2π3.5t (0.044 cos 22t) or x = 0.044 sin 2π3.5t etc
ecf for A
B1
(1)
(iii)
αmax = (2π3.5)2 0.044
C1
21 (21.3) m s–2 ecf for A and incorrect 2πf from (ii)
(0.042 gives 20.3; 0.04 gives 19.4)
A1
(2)
[6]
6
7
C
[1]
A
[1]
Page 30 of 50
8
(a)
(i)
acceleration (not a) and displacement (not x) are in
opposite directions OR restoring force/acceleration
always acts toward rest position
B1
1
(ii)
(+) sine curve consistent with a graph
B1
1
(b)
(i)
statement that EK = EP
B1
statement of max values considered
B1
EP = ½ k(Δl)2 or EPmax = ½ kA2
B1
correctly substituted values
B1
EK = 3.7 × 10–2 J
B1
OR
f = 1/T or T = 3.97 s or period equation
B1
leading to f = 0.252 Hz
B1
ωmax
= 1.58 rad s–1 or vmax = 0.055ms–1 (seen or used)
B1
substituted values into EK = ½mA2ω2 or EK = ½mv2
B1
EK = 3.7 × 10–2 J
B1
5
Page 31 of 50
(ii)
any attenuation from t = 0 seen
M1
10 mJ or E0/4 at either 4s or third hump
M1
consistent period values minima at 1 and 3s
maxima at 0 and 4s
A1
3
[10]
9
10
B
[1]
(a)
acceleration/force is directed toward
a (fixed) point/the centre/the equilibrium position
or
a = –kx + ‘–’ means that a is opposite direction to x
B1
acceleration/force is proportional to the distance from the
point/displacement
or
a = –kx where a = acceleration; x = displacement and
k is constant
B1
2
Page 32 of 50
(b)
(i)
3.2 = 2π√l/9.8 (condone use of g = 10 m s-2 for C mark)
(use of a = –ω2x is a PE so no marks)
C1
2.5(4) m
A1
2
(ii)
Correct value at 0.5 m and correct curvature
M1
Energy at 1 m = 160 J
A1
2
[6]
11
12
13
14
15
16
D
[1]
C
[1]
C
[1]
C
[2]
C
[1]
D
[1]
Page 33 of 50
17
18
19
20
21
22
A
[1]
B
[1]
C
[1]
D
[1]
B
[1]
(a)
acceleration is proportional to displacement (1)
acceleration is in opposite direction to displacement, or
towards a fixed point, or towards the centre of oscillation (1)
2
(b)
= 1.1 Hz (or s–1) (1)
(i)
f=
(ii)
(use of a = (2πf)2A gives)
(1.09 Hz)
a = (2π × 1.09)2 × 76 × 10–3 (1)
= 3.6 m s–2 (1)
(3.56 m s–2)
(use of f = 1.1 Hz gives a = 3.63 m s–2)
(allow C.E. for incorrect value of f from (i))
Page 34 of 50
(iii)
(use of x = A cos(2πft) gives)
x = 76 × 10–3 cos(2π × 1.09 × 0.60) (1)
= (–)4.3(1) × 10–2m (1)
(use of f = 1.1 Hz gives
(43 mm)
x = (–)4.0(7) × 10–2 m
(41 mm))
direction: above equilibrium position or upwards (1)
6
(c)
(i)
graph to show:
correct shape, i.e. cos curve (1)
correct phase i.e. –(cos) (1)
(ii)
graph to show:
two cycles per oscillation (1)
correct shape (even if phase is wrong) (1)
correct starting point (i.e. full amplitude) (1)
max 4
[12]
23
24
25
26
C
[1]
A
[1]
C
[1]
C
[1]
Page 35 of 50
27
28
29
30
31
C
[1]
D
[1]
B
[1]
D
[1]
(a)
(i)
speed at P, v (=
)=
= 22(.1) (m s–1)
2
(ii)
use of F = k∆L gives
= 11 (10.5) (m)
2
(b)
(i)
period T = 2π
= 2π
(= 6.51 s)
time for one half oscillation = 3.3 (3.26) (s)
2
Page 36 of 50
(ii)
(= 0.154 (Hz))
frequency
use of v = ±2πf
2
when x = 10.5 m and v = 22.1 m s–1 gives 22.12
0.1542 (A2 – 10.52)
= 4π × >
from which A = 25.1 (m)
[alternatively, using energy approach gives ½ mvP2 + mg∆L = ½ k(∆L)2
∴ (29 × 22.12) + (58 × 9.81 × ∆L) = 27 (∆L)2
solution of this quadratic equation gives ∆L = 35.7 (m)
from which A = 25.2 (m)
]
3
(c)
bungee cord becomes slack
student’s motion is under gravity (until she returns to P)
has constant downwards acceleration or acceleration is not ∝ displacement
2
(d)
(i)
when student is at R or at bottom of oscillation
1
(ii)
at uppermost point or where it is attached to the railing
because stress = F/A and force at this point includes weight of whole cord
[accept alternative answers referring to mid-point of cord because cord
will show thinning there as it stretches or near knots at top or bottom
of cord where A will be smaller with a reference to stress = F/A]
2
[14]
32
33
34
B
[1]
A
[1]
(a)
(i)
elastic potential energy and gravitational potential energy ✓
For elastic pe allow “pe due to tension”, or “strain energy” etc.
1
Page 37 of 50
(ii)
elastic pe → kinetic energy → gravitational pe
→ kinetic energy → elastic pe ✓✓
[or pe→ke→pe→ke→pe is ✓ only]
[or elastic pe → kinetic energy → gravitational pe is ✓ only]
If kinetic energy is not mentioned, no marks.
Types of potential energy must be identified for full credit.
2
(b)
(i)
period = 0.80 s ✓
during one oscillation there are two energy transfer cycles
(or elastic pe→ke→gravitational pe→ke→elastic pe in 1 cycle)
or there are two potential energy maxima per complete oscillation ✓
Mark sequentially.
2
(ii)
sinusoidal curve of period 0.80 s ✓
– cosine curve starting at t = 0 continuing to t = 1.2s ✓
For 1st mark allow ECF from T value given in (i).
2
(c)
(i)
use of T =
∴
gives 0.80 =
= 22 (21.6) ✓
✓
N m–1 ✓
Unit mark is independent: insist on N m–1.
Allow ECF from wrong T value from (i): use of 0.40s gives
86.4 (N m–1).
3
Page 38 of 50
(ii)
maximum ke = ( ½ mvmax2) = 2.0 × 10–2 gives
vmax2 =
✓ (= 0.114 m2s–2) and vmax = 0.338 (m s–1) ✓
✓
vmax = 2πfA gives A =
and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm
[or maximum ke = (½ mvmax2) = ½ m (2πfA)2 ✓
½ × 0.35 × 4π2 × 1.252 × A2 = 2.0 × 10–2 ✓
∴ A2 =
✓ ( = 1.85 × 10–3 )
and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]
[or maximum ke = maximum pe = 2.0 × 10–2 (J)
maximum pe = ½ k A2 ✓
∴ 2.0 × 10–2 = ½ × 21.6 × A2 ✓
from which A2 =
✓ ( = 1.85 × 10–3 )
and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]
First two schemes include recognition that f = 1 / T
i.e. f = 1 / 0.80 = 1.25 (Hz).
Allow ECF from wrong T value from (i) – 0.40s
gives A = 2.15 × 10–2m but mark to max 3.
Allow ECF from wrong k value from (i) –86.4Nm–1 gives
A = 2.15 × 10–2m but mark to max 3.
4
[14]
35
D
[1]
Page 39 of 50
36
(a)
(i)
correct period read from graph or use of f=1/T 0.84±0.01
C1
2.4 Hz gets C1
correct frequency 1.2 (1.18 − 1.25 to 3 sf)
A1
(ii)
correct shape (inverse)
B1
Crossover PE = KE
B1
(b)
(i)
C1
48.7 (49) m
A1
(ii)
v = 120 000 / 3600 = 33(.3) m s−1
B1
Use of F = m v2/r (allow v in km h−1 )
B1
Total tension = 6337 + (280 × 9.81) = 9.083 × 103 N
Allow their central force
B1
Divide by 4
2.27 × 103 N
Allow their central force
B1
(iii)
mgh = ½ mv2
B1
Condone: Use of v = 2πfA (max2)
9.8×44 = 0.5 v2
Allow 45 in substitution
B1
Page 40 of 50
Condone
29.4 m s−1
22 m s−1
(Use of 45 gives 29.7)
B1
106 km h−1 (their m s−1 correctly converted)
Or compares with 33 m s−1
B1
(iv)
1/16th(0.625) % of KE left if correct
M1
Allow 1/8 (0.125)or 1/32(0.313)
KE at start = 5.6 × 104 J or states energy ∝ speed2 so speed is ¼
M1
Allow for correct subn E =½ 280 × 202 x factor from incorrect
number of swings calculated correctly
Final speed calculated = 5 m s−1
A1
Must be from correct working
[17]
37
(a)
acceleration is proportional to displacement (from equilibrium)
Acceleration proportional to negative displacement is 1st mark only.
acceleration is in opposite direction to displacement
or towards a fixed point / equilibrium
Don’t accept “restoring force” for accln.
position
2
Page 41 of 50
(b)
(i)
3SF is an independent mark.
When g = 9.81 is used, allow either 0.502 or 0.503 for 2nd and 3rd
marks.
]
Use of g = 9.8 gives 0.502 Hz: award only 1 of first 2 marks if
quoted as 0.502, 0.503 0.50 or 0.5 Hz.
answer to 3SF
3
(ii)
Allow ECF from any incorrect f from (b)(i).
= 0.42 (0.419) (m s−2)
2
(c)
recognition of 20 oscillations of (shorter) pendulum
and / or 19 oscillations of (longer) pendulum
Explanation: difference of 1 oscillation or phase change of 2
or Δt = 0.1 so n = 2 / 0.1 =20, or other acceptable point
time to next in phase condition = 38 (s)
Allow “back in phase (for the first time)” as a valid explanation.
[ or (T = 1.90 s so) (n + 1) × 1.90 = n × 2.00
gives n = 19 (oscillations of longer pendulum)
minimum time between in phase condition = 19 × 2.00 = 38 (s)
]
3
[10]
38
39
40
D
[1]
D
[1]
C
[1]
Page 42 of 50
Examiner reports
1
2
3
5
This question on simple harmonic motion, readily gave the correct answer to students who could
apply a = −(2πf)2 x : clearly therefore amax is proportional to xmax, the amplitude. 78% of the
responses were correct. Incorrect answers were fairly evenly distributed amongst the other three
distracters.
This question required the selection of a correct acceleration-displacement graph for a body
moving with shm. 70% chose the correct one – a straight line of negative gradient – but 19%
opted for the parabolic shape shown by distractor A. The facility of this question improved by
25% since it was used in a 2011 examination.
This question tested knowledge of acceleration in SHM and was answered correctly by almost
two-thirds of the candidates, which compares favourably with two-fifths in a previous AS
examination.
(a)
The majority of the candidates gained full credit for this question. The most common cause
of a lost mark was failure to show clearly that the total energy was constant. This was
judged by showing that KE and PE were equal at 1 × 10–2 J.
(b)
(i)
Misreading the amplitude as 0.042 m or giving the maximum energy, 2 × 10–2 J were
common errors.
(ii)
There were many correct responses. There was a significant proportion, however,
who did not understand the question and gave the formula for the period of the
pendulum.
(iii)
Allowing for errors carried forward this part was well done by most candidates. Giving
the unit as m s–1 was a commonly seen error.
8
(a)
(i)
The clear majority of the candidates were able to explain that the minus sign showed
either that the acceleration was directed towards the equilibrium position or else that
it indicated that the acceleration and the displacement were in opposite directions.
Weaker candidates suggested that it meant that there was always deceleration or
that the acceleration was in the opposite direction to the ‘motion’.
(iii)
Most candidates recognised that the velocity would be a positive sine curve, although
other sinusoidal curves were all relatively common.
Page 43 of 50
(b)
10
(i)
The two routes to solution of this part were equally common. Those candidates opting
for the ‘elastic potential energy’ version often failed to explain that in equating the
elastic potential energy to the kinetic energy, they were using the maximum values.
Those candidates calculating the kinetic energy through the frequency or angular
frequency, often missed out the intermediate steps that a ‘show that’ equation
requires. This second method gave the period that was needed to answer part (b)(ii).
(ii)
The majority of candidates found this part to be difficult, with very few scoring all three
marks. Although virtually all candidates attempting this part were credited for showing
attenuation of the signal, most did not recognise that after one period the energy
would have fallen to a quarter of the maximum value when the amplitude fell to half
its maximum value. Very few candidates drew an accurate curve showing this.
(a)
The majority appreciated what was required but many failed to express one or both
conditions clearly. Confused candidates referred to acceleration being proportional to and
in the opposite direction to velocity or amplitude. A significant proportion stated only that for
SHM the period is independent of amplitude which gained no marks.
(b)
(i)
This was generally well done. Some inappropriately used the SHM equation
g = –(2π/3.2)2x in which x was equal to the length of the string. This gave the
correct numerical value but the use of wrong physics resulted in no marks.
(ii)
14
15
16
There were few correct answers to this part. Many who were on the right lines failed
to get the value at 1.00 m correct. There were many who simply drew a straight line
through the origin. Careless reading led many to decide that this question was about
the total energy remaining constant during one cycle of SHM. Others drew graphs
showing the variation of KE or PE (or both) with displacement.
This question, on simple harmonic motion, required candidates to substitute the given values in
the equation v2 = (2πf)2(A2 – x2), as well as to appreciate that f = 1/T. This caused fewer problems
than expected, because the facility rose to 80% from a facility of 69% when it was last used in a
linear A level examination.
This question amounted to a two-stage calculation on simple harmonic motion. The facility of
84% was a significant advance on that of 67% in the pre-examination test. Wrong responses
were almost equally divided between distractors A, B and D.
This question tested the graphical relationship between kinetic energy and displacement in
simple harmonic motion. The facility of 59% was an improvement over the 50% achieved when
this question was pre-tested. Almost one in five of the candidates chose distractor A, forgetting
that there are two cycles of energy for every cycle of displacement.
Page 44 of 50
17
18
19
20
21
This question tested candidates’ understanding of the acceleration of a particle moving with
simple harmonic motion. Over half of the candidates gave the correct response, but one in five of
them thought that the acceleration was greatest at zero displacement.
This question was concerned with the phase difference between velocity and displacement in
simple harmonic motion. The facility of 59% corresponded exactly with that in the preexamination
test. Candidates who chose wrong answers were almost equally divided between distractors A
and D, suggesting that there is much confusion in understanding whether 90° means ð/2, ð/4, or
2ð radians.
Simple harmonic oscillation was the topic tested by this question (which had been used before).
79% of the responses were correct, which was 8% better than when it was last used. This shows
that vmax = 2πfA is well known.
In this question candidates had to know how the acceleration-displacement relationship for a
simple harmonic oscillator translates into a graph; linear but having a negative gradient. Slightly
less than half of the candidates chose the correct response. Almost as many chose distractor A,
which would be the potential energy-displacement graph.
This question, about phase differences in shm, had a facility of 72% but did not discriminate very
well. 17% chose distractor A (π/4 instead of π/2); this may have been caused by a
misunderstanding of the radian to degree conversion.
Page 45 of 50
22
The conditions expected in answers to part (a) were those embodied in the definition of shm:
acceleration is proportional to displacement, but acts in the opposite direction (or towards a fixed
point / towards equilibrium position). Other features of the acceleration, such as the fact that a is
a maximum when v = 0, were not given any credit.
In part (b) frequency was often confused with period; of itself this was only penalised once,
leaving five marks available. Part (b) (iii) caused problems for many candidates, mainly because
they did not realise that 2πf t is an angle measured in radians rather than degrees. Several
candidates confused acceleration a and amplitude A, leading to the incorrect substitution x =
3.56 cos (2π × 1.09 × 0.60). Another prevalent wrong answer was 49 or 50 mm, apparently
arrived at by calculating (0.60/0.92) × 76 mm, i.e. ( t / T ) × A. The direction of displacement when t
= 0.60s could be arrived at heuristically, without resort to the result of the previous calculation;
the direction mark was therefore regarded as independent.
Many very good answers were seen for the graphs in part (c). Common errors were a cos graph
(rather than – cos) in (i), and the wrong shape of Ep curve – even when it had been appreciated
that there are two energy cycles per oscillation – in (ii).
23
24
25
The outwardly more demanding this question, which had appeared in a previous PA04 test, was
also on shm. It required an algebraic expression for the magnitude of the acceleration of a body
when at maximum displacement. This time 75% of the candidates gave the correct response, but
perhaps it was the same 11% of them that were tempted by distractor A (zero acceleration). This
was the most discriminating question on the paper.
In this question, 64% of the responses were for the correct shape of Ek against x curve. By
selecting distractor B, 25% of the candidates showed that they understood the trend of the Ek
cbehaviour but not the exact form of it.
This question was concerned with simple harmonic motion. Candidates found the question,
which had appeared in a previous examination, slightly easier than last time; three-quarters of
them successfully applied a = (2πf)2 A to give the correct answer. This was the most
discriminating question in the examination.
26
This question does not appear to be easy – albeit it that a correct answer only requires care
when working out the result from three values substituted into v = ± 2πf (A2 – x2). Yet, as on the
previous occasion when this question was used, it turned out to have the highest facility on the
paper – it was 88% this time.
Page 46 of 50
27
28
29
This question was another easy question, the correct kinetic energy-time graph being recognised
by over 80% of the candidates.
This question involved a two-stage calculation on the connection between the energy of a
particle moving with simple harmonic motion, its mass and the time period. This proved to be
demanding, as it had been when the question last appeared in an examination. On the previous
occasion slightly fewer than half of the candidates gave the correct answer, whereas this time it
was slightly more than half. The question discriminated between the abilities of candidates better
than any other in this test. Perhaps it was an inability to translate from ω = 1 to T = 2π/1 that
caused almost a quarter of the candidates to select distractor B, where the answer was π / 2
instead of 2π.
This question required candidates to calculate the maximum acceleration of an object moving
with shm by reading its maximum displacement and time period from a displacement-time graph,
and then applying amax = (2πf) 2 xo. Three quarters of them were successful in this.
30
This question also re-banked questions, were about features of simple harmonic motion. Their
facilities were 67% and 61% respectively, both significant improvements on the previous
results.Those who chose distractor D in Question 7 (22% of students) clearly realised that the
graph of kinetic energy against distance should be curved, but they chose the wrong shape of
curve.
Page 47 of 50
31
This question, based on bungee jumping, tested simple harmonic motion in an unfamiliar context
and at the same time to provide a synoptic test of some AS content. Examiners were pleased to
see that a high proportion of the students were able to cope competently with this unfamiliar
situation.
Application of energy conservation, or of the equations for uniformly accelerated motion under
gravity, led to a high proportion of correct answers in part (a) (i). The equation representing
Hooke’s law was well known in part (a) (ii) but a few students showed confusion between mass
and weight.
Part (b) (i), which required the time for half of an oscillation, only caused problems for the small
number of students who misinterpreted the wording and determined the time for one-and-a-half
oscillations. Part (b) (ii) was much more challenging and turned out to be a question that many
students returned to answer on a supplementary sheet. The most direct solution came by
applying the equation
, with careful choice of the earlier values obtained
for v and x, and of the derived value for f. Most students seemed to think a quick solution could
be arrived at by applying vmax = 2πfA, but this is incorrect. It is possible to reach a correct solution
from energy considerations; this needs particular care over the balance of gravitational pe lost,
ke gained and elastic pe gained at some consistent point in the motion. Nevertheless, a few
correct solutions using this approach were seen.
In part (c) most students realised that the bungee cord would cease to exert a force on the
bungee jumper once she was higher than point P. Few went on to mention that her motion was
then purely under gravity or that her acceleration became constant, although references to the
fact that acceleration would no longer be proportional to displacement were quite common.
Almost all students gave the correct answer – point R – in part (d) (i). The responses in part (d)
(ii) revealed a widespread misunderstanding of the significance of centre of mass, with
statements such as ‘the stress is a maximum at the centre of the cord because that is where the
weight acts’ seen. Acceptable answers included at the point where the cord is attached to the
railing (where the greatest weight is supported) and (because of possible thinning) half way along
the cord. It was expected that students would show that they understood what is meant by stress
when formulating their reason, whichever point in the cord they gave.
32
33
This question had also appeared in a previous examination but the improvement in its facility
(62% to 77%) was less spectacular. Those who could combine ½ mv2 with vmax = 2πfA, deduce
that f would be 50 / 47, and then substitute the other given values correctly would easily arrive at
the correct value for maximum kinetic energy.
This question also re-banked questions, were about features of simple harmonic motion. Their
facilities were 67% and 61% respectively, both significant improvements on the previous
results.Those who chose distractor D (22% of students) clearly realised that the graph of kinetic
energy against distance should be curved, but they chose the wrong shape of curve.
Page 48 of 50
34
Part (a)(i) offered an easy mark for naming the two types of potential energy involved in an
oscillating suspended mass-spring system. “Gravitational potential energy” is clear and
unambiguous, but a variety of terms appeared to be in use for the energy stored by a stretched
spring. “Elastic potential energy” was the expected term, but “strain energy” was equally
acceptable. For obvious reasons “stored energy” (when unqualified) was not.
Those who had concentrated on the wording of the question in part (a)(ii) – especially on “energy
changes”, “one complete oscillation” and “starting at its lowest point” – were able to give good
concise answers. Far too many of the students attempted to consider the absolute values of
elastic and gravitational energies during an oscillation, which usually led them into confusion and
irrelevance. Many answers stated that elastic potential energy would increase as the mass
moved above the equilibrium position because the spring would be compressed. Inevitably a lot
of answers described only half of an oscillation: if the energy types and changes were correctly
described even this was given 1 mark. Answers which did not refer to the kinetic energy of the
system were not credited.
In part (b)(i) those who appreciated that the total potential energy of the system passes through
two maxima per oscillation, one at each amplitude, came up with the expected 0.8 s. Because
they did not understand this, getting on for half of the students gave 0.4 s. There were
consequences in the later parts of this question, where incorrect values from part (b)(i) were
generally accepted as a basis for the work that followed. A small minority of the graphs drawn in
part (b)(ii) were triangular, but the majority represented some form of sinusoidal variation.
Whether this agreed with the expected period (0.8 s), and was a negative cosine curve, proved
to be more testing issues.
The time period calculation in part (c)(i) was straightforward. This was rewarding for those who
could substitute mass and period values correctly and then calculate the expected value. N m−1
was the only answer accepted for the unit of k. The amplitude calculation in the final part of the
question was often done well. Students who had made an error over the time period earlier were
unable to show that the value of the amplitude would be about 40 mm, so were limited to a mark
of 3 out of 4. Working from T = 0.4 s, many of these answers arrived at a value of 21.5 mm
before the students introduced a mystery factor of 2 to end up with “about 40 mm”!
35
This question on simple harmonic motion was one of the most demanding questions in the test.
Candidates should have realised that the total energy E of the oscillating particle is equivalent to
its maximum kinetic energy ½ m vmax2, and that vmax= 2πfA. When the amplitude A is constant it
follows that E is proportional to mf2. Doubling both m and f therefore produces an eightfold
increase in E. Only 42% of the candidates selected the correct response. Almost inevitably,
distractor C – a fourfold increase in E – was the second most popular choice with 32% of the
responses. This question was also a good discriminator.
Page 49 of 50
36
(a)
(b)
(i)
A minority of the candidates realised that the PE reaches a maximum twice per cycle
so identifying a period of 0.4 s was most common.
(ii)
Most drew diagrams that showed the KE at peak when the PE was at zero. The
majority however either did not know or were careless in showing that PE = KE when
the graphs cross over.
(i)
This was a straightforward question which produced many correct answers. Poor
algebra was a problem for those who were unsuccessful.
(ii)
Most were able to calculate the speed in m su−1 and go on to find the centripetal
force. Fewer realised that the weight of the riders and harness had to be added to
this to find the sum of the tensions in the cables. Many however did remember to
divide what they thought was the tension by 4.
(iii)
Determining the number of half oscillations proved difficult for some candidates but
most appreciated the need to do a (0.5)n type calculation to find the final energy. A
common error was to multiply the fractional change (1 / 16) by the initial speed of 20
m s−1.
37
Except for part (c), this question was on material familiar to most candidates, so the marks
awarded were generally high. In part (a) definitions of simple harmonic motion, in terms of the
two features of the acceleration of a body moving with shm, were generally well known. Part
(b)(i) presented a greater challenge for candidates who were unsure about how to handle
significant figures. All the data required to complete the calculation in this question was available
to 3SF. Therefore all the working should have been to at least 3SF and the answer should be
quoted to 3SF. Candidates who used g = 9.8 instead of g = 9.81 lost a mark, and a further mark
was lost if the final answer was not expressed to 3SF. Some of the weaker candidates did not
appear to know the difference between period and frequency. Part (b)(ii) required a
straightforward application of a = (–) (2πf)2 × and presented few problems.
The solution of the question in part (c), involving the minimum time period between ‘in phase’
positions of two pendulums of different frequencies, could not be arrived at by a standard method
that most candidates would have encountered. Consequently the explanations of how the
answer had been arrived at were often unsatisfactory. Trial and error seemed to be a popular
approach, sometimes leading to the correct answer of 38s without any working at all.
Probably the most satisfactory solution is to recognise that the shorter pendulum must make one
more oscillation than the longer pendulum in the required time, hence the number of oscillations
of the longer pendulum is given by (n + 1) × 1.9 = n × 2.0. Another successful approach follows
from appreciating that Δt = 0.1s, so the number of oscillations required of the shorter pendulum is
2.0 / 0.1 = 20 (alternatively 1.9 / 0.1 gives 19 oscillations of the longer pendulum). This approach
sometimes led to an incorrect conclusion, such as 19 × 1.9 = 36.1s, or 20 × 2.0 = 40s.
40
This question concerned the equations of simple harmonic motion. In this question, the selection
of the correct algebraic terms for amplitude and frequency from the equation x = P cos Qt was
carried out correctly by three-quarters of the students. Difficulties with algebra are likely to have
caused 17% to select distractor B (where frequency = 2π Q instead of Q / 2π).
Page 50 of 50