93/11(a)
Semester 1, 2009
THE UNIVERSITY OF SYDNEY
PHYS 1901
PHYSICS 1A (ADVANCED)
Solutions
JUNE 2010
Time allowed: THREE Hours
MARKS FOR QUESTIONS ARE AS INDICATED
TOTAL: 90 MARKS
INSTRUCTIONS
•
•
•
All questions are to be answered.
Use a separate answer book for section A and section B.
All answers should include explanations in terms of physical principles.
DATA
ρ
=
1.000 ×103 kg.m −3
Free fall acceleration at earth's surface g
=
9.80 m.s −2
Gravitational constant
G
=
6.67 ×10−11 N.m 2 .kg −2
Speed of light in a vacuum
c
=
3.00 ×108 m.s −1
Speed of sound in air
v
=
344 m.s −1
NA
=
6.023 ×1023 mol−1
Universal gas constant
R
=
8.314 J.mol−1.K −1
Boltzmann constant
k
=
1.380 ×10−23 J.K −1
Stefan-Boltzmann constant
σ
=
5.67 ×10−8 W.m −2 .K −4
Density of fresh water
Avogadro constant
Page 1 of 1
ADV_Q01=REG_Q01
Question 1
A mass of 2m is connected to two masses of mass m and 3m as shown below. The strings
connecting the masses are light. The pulleys and the slopes are frictionless.
(a)
Draw separate free body diagrams for each of the masses, taking care to identify all forces
acting.
(b)
If the angles of the slopes are θ = 30 D, and the system is released from rest, does the
hanging mass 2m remain where it is, accelerate upwards or accelerate downwards? Justify
your answer.
(5 marks)
Solution
(a)
(1 mark for left diagram; ½ mark each for similar right diagrams)
(b)
Take vertically upwards as the positive direction for the 2m mass and downwards along the slope
as the positive direction for the m and 3m masses
Apply Newton’s Second Law to the 2m mass as follows:
T1 + T2 − 2 m g = 2m a
⇒ T1 + T2 = 2m a + 2 m g
Apply Newton’s Second Law to the m mass as follows:
T1 − m g sin θ = − m a
⇒ T1 = m g sin θ − m a
Apply Newton’s Second Law to the 3m mass as follows:
T2 − 3m g sin θ = − 3m a
⇒ T2 = 3m g sin θ − 3m a
Equation (1) – Equation (2) – Equation (3)
(T1 + T2 ) − T1 − T2
= (−2 m a + 2 m g ) − (−ma − mg sin θ ) − (3 m a − 3 m g sin θ )
−6 m a + 2 m g − 4 m g sin θ
=0
⇒ − 6 m a = 2 m g − 4 m g sin θ
⇒ a = ( g − 2 g sin θ ) / 3
For θ = 30D we have sin θ = 0.5
(1)
(2)
(3)
a = ( g − 2 g (0.5)) / 3 = 0 .
So the system is in equilibrium.
Alternate Derivation
With insight, assume system is equilibrium and calculate tensions T1 and T2 finding that
4 m g sin θ = 2 m g
(1 mark for answer; 2 marks for valid justification)
ADV_Q02
Question 2
(a)
(b)
A very heavy 4WD collides head-on with a very light Mini Cooper. For each of the
following claims, state whether it is correct or incorrect and briefly justify your answer.
(i)
The amount of kinetic energy lost by the 4WD is equal to the amount of kinetic
energy gained by the Mini.
(ii)
The change of momentum of the 4WD is equal to the change of momentum of the
Mini.
(iii)
The Mini experiences a considerably greater force during the collision than the
4WD.
Should the design of a car aim to make a collision with another vehicle as elastic or
inelastic as possible? Explain your answer.
(5 marks)
Solution
(a)
(i) Incorrect. Kinetic energy is not conserved in the collision, which is inelastic.
(1 mark)
(ii) Correct. All forces are internal to the system during the collision. Total momentum is
therefore conserved in the collision.
(1 mark)
(iii) Incorrect. By Newton’s 3rd law, the force on the Mini due to the 4WD is equal and opposite
to the force on the 4WD due to the Mini.
(1 mark)
(b)
To allow the passengers to survive the collision it is necessary increase the interaction time
between the cars. This minimizes the force on the car (and the passengers) because
dp
dv
F=
=m .
dt
dt
This is most easily done by making the collision as inelastic as possible so that the structure of
the car absorbs as much of the energy of the collision as possible.
A long enough interaction time could also result from a perfectly elastic collision in which the
cars (with rather impractical spring bumper bars) are brought slowly to rest and then allowed to
slowly rebound. In this case, there is a greater change of momentum than the inelastic situation
as the car changes from its initial velocity, through zero, to a rebound velocity.
(2 marks)
ADV_Q03
Question 3
The most efficient way to send a spacecraft to another planet in the solar system is by using a
Hohmann transfer orbit. If the orbits of the departure and destination planets are circular, the
Hohmann transfer orbit is an elliptical orbit whose perihelion (closest point to the sun) and
aphelion (furthest point from the sun) are tangent to the orbits of the two planets (see diagram
above).
(a)
The rockets are fired briefly at the departure planet (say, Earth) to put the spacecraft into
the transfer orbit. To move into an orbit with a larger radius (such as Mars), should the
rockets be fired in the direction of motion, or opposite to the direction of motion?
Explain your answer.
(b)
The spacecraft then coasts (does not use its rockets) until it reaches the destination planet.
Does the speed of the spacecraft increase, decrease, or remain constant during this stage?
Explain your answer.
(c)
At the final stage, the rockets are again fired at the destination planet (Mars) to match its
orbit. Should the rockets be fired in the direction of motion, or opposite the direction of
motion? Explain your answer.
(d)
Ignoring the gravitational potential energy due to the planets and using only that due to
the sun (mass M = 1.99 × 1030 kg ), what is the change in energy (potential energy +
kinetic energy) of a spaceship of mass 1.00 × 106 kg having moved from a circular orbit
of rE = 1.50 × 1011 m (Earth’s orbit) to a circular orbit of rM = 2.28 × 1011 m (Mars’ orbit)?
Explain why the energy of the spacecraft has increased/decreased in terms of work done
by its rockets.
(5 marks)
Solution
(a)
Rockets must be fired in the direction of motion.
(½ mark)
From a circular orbit, the spacecraft must increase its speed to a value greater than that required
to maintain circular orbit. Therefore, the rockets must be fired such as to increase the speed.
This does work in the direction of motion and therefore increases the spacecraft energy.
(½ mark)
(b)
The spacecraft’s speed decreases.
(½ mark)
GM m
increases (becomes
r
less negative). Because it is coasting, total energy is conserved, and therefore the kinetic energy
must decrease.
(½ mark)
As the radius is increased, the gravitational potential energy U = −
(c)
Rockets must be fired in the direction of motion.
(½ mark)
From an elliptical orbit at the aphelion, the spacecraft is moving too slow relative to an object in
a circular orbit with the same radius. Therefore, the rockets must be fired such as to increase the
speed by doing work in the direction of motion.
(½ mark)
(d)
The total energy is
E (r ) = U (r ) + K (r )
where
GM m
U (r ) = −
r
and
GM m
K (r ) =
2r
Giving
GM m
E (r ) = −
.
2r
The change in energy is
ΔE = E (rM ) − E (rE ) = −
GM m⎛ 1 1 ⎞
⎜ − ⎟
2 ⎝ rM rE ⎠
= 1.51×1014 J .
(1 mark)
Its total energy increases, because the rockets have done positive work on the spacecraft at both
stages.
(1 mark)
ADV_Q04=REG_Q04
Question 4
(a)
Before giving you an injection, a physician swabs your arm with isopropyl alcohol, a
volatile liquid at room temperature.
(i)
Why does this make your arm feel cold?
Because of a fear of needles, you sit on a chair gripping the metal armrests with your
hands.
(ii)
Why do the metal armrests feel cool to the touch?
(b)
Use the concepts of the kinetic-molecular model to explain why the pressure of an ideal
gas in a container increases as heat is added to it.
(c)
Explain in terms of entropy and the Second Law of Thermodynamics why air molecules
in a sealed room do not spontaneously collect into one half of the room, occupying half
the volume and leaving a vacuum in the other half. Your answer should include
consideration of the number of microscopic states of the molecules in the room.
(5 marks)
Solution
(a)
(i)
Evaporative cooling – To evaporate the alcohol requires heat (its latent heat of
vaporization). This comes from the person’s skin leaving it cooler than before.
(1 mark)
(ii)
Conduction – metal is a good thermal conductor, so it effectively transfers away the heat
from your hands.
(1 mark).
(b)
3
1
k T = K p = m v 2 avg which is the average translational kinetic energy of the
2
2
gas molecules. When heat is added to the gas, the average molecular speed increases and so does
T and p.
(1 mark)
pV = N k T and
(c)
Answer 1
The entropy statement of the Second Law says that the entropy of a closed system can never
decrease, so a closed system can never spontaneously undergo a process that decreases the
number of possible microstates.
S = k ln w
where, S is entropy and, w is the number of possible microstates. The Second Law states that
ΔS ≥ 0 and so the number of microstates cannot decrease (which would be the case if all
molecules were in one half of the volume).
Answer 2
The Second Law states that the entropy or degree of disorder of a closed system increases or
remains constant. Putting all molecules into a half of the volume of the room corresponds to
greater order (i.e. decreased disorder). It is therefore inconsistent with The Second Law.
Answer 3
If N air molecules spontaneously collected in half the volume of the room, then the number of
possible microstates would decrease by a factor 2 N . The probability of finding a given molecule
in one half of the room is ½ so the probability of finding all N molecules in half the room is
N
⎛1⎞
⎜ ⎟ which for a large number of molecules is very small.
⎝2⎠
(2 marks for reasonable explanation)
ADV_Q05
Question 5
An object with mass m moves in one dimension under the influence of a force given by
F ( x) = −k x − β x3 ,
where x is its position.
(a)
The object will oscillate about the position x=0. Briefly explain how we know this.
(b)
Does the object undergo simple harmonic motion? Explain briefly.
(c)
Assuming the amplitude of the oscillation is very small, estimate the period of the
oscillation, giving reasons for your answer.
(5 marks)
Solution
(a)
The force is in the opposite direction to the displacement (minus sign in the equation) and is zero
at the origin, so it acts to restore the object to its central position.
(2 marks)
(b)
No, SHM only occurs if the force increases linearly with displacement.
(1 mark)
(c)
For small amplitudes the linear term in the force equation will dominate and motion will be
approximately simple harmonic. So the period will be the same as for F = − k x , namely
T = 2π
m
.
k
(2 marks)
ADV_Q06
Question 6
Consider three objects in gravitational orbits around each other, such as three stars or two stars
and one planet. In some circumstances, such a system can exhibit chaotic behaviour.
(a)
Explain briefly what is meant by the statement that the system exhibits chaotic behaviour.
(b)
In class we learned that for a system to exhibit chaos, there must be some non-linearity in
the system. What is the non-linearity in this system?
(5 marks)
Solution
(a)
The motion is irregular, not periodic, and a slight change in the initial conditions results in a
change in subsequent behaviour that diverges exponentially. The motion cannot be predicted
from the initial conditions.
(3 marks)
(b)
The force of gravity is an inverse square law, so the attractive forces between the objects varies
non-linearly with their separations.
(2 marks)
ADV_Q07
Question 7
A steamroller of total mass M is constructed so that all of its mass is in two rollers, which are
solid cylinders of uniform density and radius R . The centres of the rollers are separated by a
rigid, massless rod of length d = 3 R .
The masses of the cylinders are not the same. The front one has a mass M 1 = M / 3 , while the
back one has a mass M 2 = 2 M / 3 . The moment of inertia of a uniform cylinder of mass m and
radius r is
m r2
.
2
The steamroller is located on a slope of angle θ to the horizontal, as shown in the diagram.
(a)
Where is the centre of mass of the steamroller?
(b)
Suppose the steamroller’s brakes are released and it rolls downhill. Neglecting rolling
friction, show that its speed after descending a vertical distance h is
4gh
v=
.
3
(c)
Using the equation in part (b), or by other means, show that the acceleration of the
steamroller in part (b) is given by
2
a = g sin θ ,
3
directed downward along the slope.
(d)
If the brakes are kept on to hold the steamroller stationary
(i)
derive an expression for the maximum angle of slope θ max on which the
steamroller can be parked without sliding downhill.
(ii)
calculate the value of θ max if the coefficient of static friction between the rollers
and the ground is μ s = 0.50.
(10 marks)
Solution
(a)
Set up x-y coordinates as shown
Coordinates of centre of mass: yCM = 0 by symmetry.
xCM
M
2M
×0+
×d
3
3
xCM =
M 2M
+
3
3
2d
=
3
is measured from the centre of rear roller in the x direction as shown in the diagram.
(2 marks)
(b)
Work done by gravity
= Δ ( KE ) = Δ(GPE )
1
1
1
= Mv 2 + I1ω 2 + I 2ω 2
2
2
2
Both rollers have same radius so have same ω .
(1 mark)
Now ω = v / R
and
and
So Work done
1
1M 2
M1R 2 =
R
2
2 3
1 2M 2
I2 =
R
2 3
I1 =
1
1 ⎛1
1
⎞
Mv 2 + ω 2 ⎜ M R 2 + M R 2 ⎟
2
2 ⎝6
3
⎠
1
1
= Mv 2 + M v 2 (since v = Rω )
2
4
But the Work done is also equal to
⎛ M 2M ⎞
⎜ +
⎟ gh
3 ⎠
⎝ 3
This means that
v = 4 gh / 3
=
(valid derivation 2 marks)
(c)
a = acceleration in the + x direction
h
Use s (down the hill) = −
sin θ
(negative x direction) and v 2 = u 2 + 2as
Therefore
4g h
⎛ −h ⎞
= 0 + 2a ⎜
⎟
3
⎝ sin θ ⎠
−2 g
⇒a=
sin θ
3
In the downhill direction the acceleration is given by
2g
sin θ
3
or for a more effort
dv d
=
dt dt
h = x sin θ
a=
but
4 g 12
h
3
(if θ = the angle of the slope)
4g d
x sin θ
3 dt
4 g sin θ 1 dx
=
3
2 x dt
a=
Use
dx
= −v = − 4 gh / 3
dt
g sin θ
1
gh
2
3
3
h / sin θ
−2 g
=
sin θ
3
(negative sign ⇒ downhill
Then
a=
(2 marks)
(d)
Consider y components of forces
N1 + N 2 − M g cos θ = 0
(no acceleration in the y direction)
Consider x components:
F1 + F2 − M g sin θ = 0 ⇒ F1 + F2 = M g sin θ
At point of slippage say θ = θ max
F1 = μ s N1 and F2 = μ s N 2
where μ s is the coefficient of static friction
This means that
μs ( N1 + N 2 ) = Mg sin θ max
And from above
N1 + N 2 = M g cos θ max
Hence
tan θ max = μ s
θ max = tan −1 μ s = 27 D
(working 2 marks
(result 1 mark)
ADV_Q08
Question 8
A Hooke’s law force − k x and a constant conservative force F in the +x direction act on an
atomic ion.
(a)
Show that a possible potential energy function for this combination of forces is
1
F2
U ( x) = k x 2 − F x −
2
2k
(b)
Is this the only possible function? Explain your answer.
(c)
Graph U ( x) versus x for values of x between −2F / k and +4F / k . (Mark the values
of U ( x) in multiples of F 2 / k , and x in multiples of F / k .)
(d)
Identify the x values of the equilibrium positions. Are they stable or unstable?
(e)
If the total energy is E = F 2 / k , what are the maximum and minimum values of x that
the ion reaches in its motion?
(10 marks)
Solution
(a)
To be a potential energy function for this combination of forces, we must have
dU
Fnet = −
.
dx
For the given function
1
F2
U ( x) = k x 2 − F x −
2
2k
we have
dU
= kx−F
dx
Fnet = − k x + F
The net force is
(1 mark)
(1 mark)
and so
Fnet = −
dU
.
dx
(1 mark)
(b)
No, any function satisfying Fnet = −
dU
would suffice. The general solution would be U ( x ) + C
dx
for any constant C .
(1 mark)
(c)
Making the substitutions as follows
⎛ F2 ⎞
V =U /⎜
⎟
⎝ k ⎠
and
⎛F⎞
y = x/⎜ ⎟
⎝k⎠
the potential function can be written
V=
1 2
1
y − y−
2
2
This is plotted below. The units along the axes are as requested.
(2 marks)
(d)
Equilibrium occurs when
dU
Fnet = −
= 0.
dx
(1 mark)
This occurs when
Fnet = 0 = − k xeq + F
⇒ xeq =
F
k
(½ mark)
2
This is stable (because
dU
=k >0)
dx 2
(½ mark)
(e)
Maximum and minimum values of x occur when Etot = U ( xm ) at which the kinetic energy is
zero.
If we have
Etot = U ( xm ) =
F2
k
⇒ V ( y) = 1
This occurs for y = −1 and y = 3
F
3F
or
.
x = − and x =
k
k
(2 marks)
ADV_Q09
Question 9
The operation of a petrol internal combustion engine can be represented by a closed cycle A to B
to C to D and back to A. The amount of gas enclosed within one cylinder of the engine is
0.555 moles. The pressure of the gas at point A is 1.05 × 105 Pa and the volume of the gas is
1.52 ×10−2 m3 .
(a)
Find the temperature at point A.
The gas is compressed to point B by an isothermal process. The volume of the gas is reduced to
0.82 ×10−2 m3 .
(b)
What are the temperature and pressure at point B?
(c)
What is the change in internal energy of the gas U A→ B in going from A to B?
(d)
What is the work done by the gas WA→ B in going from A to B?
(e)
What is the heat flow to or from the gas QA→ B in going from A to B?
The gas then undergoes a constant volume process from B to C. The temperature at C is now
547 K . The gas then undergoes a constant pressure change from C to D and then another
constant volume change from D back to A.
(f)
Draw a neat p-V diagram (not to scale) for the cycle A→B→C→D→A. Indicate on the
diagram the net work done by the engine during one cycle.
(10 marks)
Solution
(a)
p A VA (1.05 ×105 )(1.52 × 10−2 )
=
pV = nRT ⇒ TA =
(0.555)(8.314)
nR
= 346 K
(1 mark)
(b)
Process is isothermal so TA = TB
We also have VB = 0.82 × 10 −2 m 3 .
So
p A VA = pB VB
⇒ pB =
p A VA (1.05 ×105 )(1.52 × 10−2 )
=
= 1.95 ×105 Pa.
−2
(0.82 × 10 )
VB
(2 marks)
(c)
Isothermal process so TA = TB ⇒ ΔT = 0 ⇒ ΔEint = 0.
(1 mark)
(d)
⎛ VB ⎞
⎛ 0.82 ×10−2 ⎞
WA→ B = n RT ln ⎜ ⎟ = (0.555)(8.314)(346) ln ⎜
−2 ⎟
⎝ 1.52 ×10 ⎠
⎝ VA ⎠
= −985 J.
(2 marks)
(e)
First law of thermodynamics gives
ΔEint = Q − W
ΔEint = 0 ⇒ Q − W = −985 J
(1 mark)
(f)
VB = VC
pC = pD
VD = VA
TC = 547 K
(2 marks)
(g)
Work done is shown by the area closed by the curve.
(1 mark)
ADV_Q10=REG_Q10
Question 10
(a)
An engineer has devised two heat engines, A and B, the detailed energy-flow diagrams of
which are shown below.
In the diagrams, TH is the temperature of a hot reservoir and TC is that of a cold
reservoir. QH and QC represent the quantities of heat absorbed and rejected by the
engine during one cycle. Check and explain whether both engines were designed by the
engineer using correct thermodynamic principles.
(b)
A man enters a sauna where the air temperature is Tair = 47 o C . His skin temperature is
Tskin = 36o C . Assuming he is completely naked and his skin surface area is A = 1.5 m 2 ,
calculate the net rate (in watts) at which the man’s skin is heated by:
(i)
Conduction. Assume that the heat is conducted to the skin through a 10 mm thick
layer of air (thermal conductivity k = 0.024 W.m −1.K −1 );
(ii)
Radiation. Assume that the emissivity of the skin is given by e = 1.0 .
(iii)
In addition to the energy transferred to his skin, the man is also generating heat by
internal metabolic processes. If the total energy reaching his skin is at a rate of
300 W, at what rate (in litres per hour) must perspiration evaporate from the
man’s skin in order to maintain a constant skin temperature? The heat of
vaporisation of water at 36 o C is 2.4 × 106 J.kg −1 .
(10 marks)
Solution
(a)
The First Law of Thermodynamics
Work Done
First Law
Consistent with First
Law
W = QH − QC
Heat Engine A
600 J
400 J
no
Heat Engine B
500 J
500 J
yes
Engine B is consistent with the First Law but Engine A is not consistent with the First Law.
The Second Law of Thermodynamics (engine efficiency)
Maximum Efficiency
Efficiency
T
W
e=
emax = 1 − C
QH
TH
Efficiency
e ≤ emax
Heat Engine A
0.60
0.50
no
Heat Engine B
0.50
0.40
no
Engine B violates the Second Law according to the maximum efficiency formula (which
incorporates the First Law in its derivation). Engine B is therefore not a viable engine.
The Second Law of Thermodynamics (entropy change)
Entropy Change
ΔStotal = ΔS hot + ΔSengine + ΔScold
=
Consistent with
Second Law
ΔS ≥ 0
Q
−QH
+0+ C
TH
TC
Heat Engine A
−1000
600
+0+
= +0.33 J.K −1
600
300
yes
Heat Engine B
−1000
500
+0+
= −0.33 J.K −1
500
300
no
According to this analysis of the Second Law, Engine B is not viable but Engine A is. However,
Engine A is not compatible with the First Law and so is not a viable engine.
(Engine A: 1 mark for analysis and 1 mark for conclusion. No need to use Second Law if
stated as not viable because of violation of First Law.)
Engine B
(1 mark for analysis, 1 mark for conclusion)
(b)(i)
Rate of heating by conduction:
H =kA
(Tair − Tskin )
L
⎛ 47 K − 36 K ⎞
= (0.024 W.m −1.K −1 )(1.5 m 2 ) ⎜
⎟
⎝ 0.01 m ⎠
= 39.6 = 40 W
(2 marks)
(ii)
Rate of radiative heating:
H = Ae σ (Tair 4 − Tskin 4 )
= (1.5 m 2 )(1.0)(5.67 × 10−8 W.m −1.K 4 )(( 320K ) − ( 309K ) )
4
4
= (1.5 m 2 )(1.0)(5.67 × 10−8 W.m −1.K 4 )(1.37 × 109 K 4 )
= 116 = 120 W
(2 marks)
(iii)
Q = m Lv ⇒
dm dQ / dt H tot
=
=
dt
Lv
Lv
(1 mark)
Rate of perspiration (mass) loss is given by:
H tot
300 J.s −1
=
= 1.25 ×10−4 kg.s −1
6
−1
Lv
2.4 × 10 J.kg
(½ mark)
Rate of perspiration (volume) loss is given by:
dV 1.25 × 10−4 m3
=
= 1.25 × 10−7 m3 .s −1
dt
1000s
(this assumes fresh water with density 1000 kg.m −3 )
This is converted to L.hr −1 as
(1.25 ×10−7 ) (1000 ) (60)(60) = 0.45 L.h −1 (litres per hour)
(½ mark)
Some students will add answers to part (i) and (ii) to the 300 W . Give full marks if otherwise
correct.
REG_Q11=ADV_Q11
Question 11
Figure A shows a 1.0 m long pipe which is closed at one end and open at the other. The air in the
pipe is vibrating as a standing wave. Figure B shows a representation of the displacement of an
air molecule relative to its equilibrium position as a function of position x along the pipe at one
instant of time.
(a)
Copy Figure B and add to it a dashed line showing the displacement at a time one half
period later.
(b)
Make a new figure in your answer book, similar to Figure B but showing the pressure
variation as a function of position x along the pipe at the same instant of time as in
Figure B.
(c)
The speed of sound in air is 344 m.s −1 . Calculate the wavelength and frequency of the
standing wave mode shown in Figure B.
(d)
What is the fundamental frequency of the pipe?
The pipe is now filled with helium (speed of sound in helium is 965m.s −1 ).
(e)
What are the wavelength and frequency for the mode shown in Figure B in this case?
(10 marks)
Solution
(a)
The curve T / 2 later is the inverted curve as shown below (dashed line)
(2 marks)
(b)
The correct pressure curve should show pressure nodes where there are displacement anti-nodes
and vice versa. The pressure varies around atmospheric pressure p0 as in the diagram below.
(1 mark for this or same but upside down)
The correct phase would have the curve being less than atmospheric pressure at x = 0 as above.
The correct phase can be seen from the positive to negative transitions on the displacement
curve. While positive the displacement of molecules is in the + x direction and when then
negative the displacement is towards the − x direction creating a higher than atmospheric
pressure around the transition point.
(1 mark for correct phase)
(c)
λ=
4m
= 0.8 m
5
(1 mark)
f =
v
λ
=
344 m.s
0.8 m
−1
= 430 Hz
(1 mark)
(d)
For fundamental
λ = 4.0 m
(1 mark)
f =
v
λ
=
344 m.s
4m
−1
= 86 Hz
(1 mark)
(e)
λ = 0.8 m
(1 mark)
f =
v
λ
=
965 m.s −1
= (1206) 1210 Hz
0.8 m
(1 mark)
ADV_Q12
Question 12
The one-dimensional wave equation is:
∂2 f
1 ∂2 f
=
.
∂ x 2 v 2 ∂t 2
Consider the specific example of a horizontal stretched string.
(a)
What physical quantity does the function f ( x, t ) represent?
(b)
What physical quantity does the symbol v represent?
Now consider a wave with wavelength λ and period T described by the function
f ( x, t ) = A cos ( k x + ω t ).
(c)
What physical quantities do the symbols ω and k represent in terms of λ and T ?
(d)
In what direction is the wave travelling? Briefly justify your answer.
(e)
Show that this function is a solution of the wave equation and hence find an expression
for v in terms of ω and k .
(f)
It turns out that the function
f ( x, t ) = A sin( k x ) cos(ω t )
is also a solution of the wave equation. What type of wave does this represent?
(10 marks)
Solution
(a)
The vertical displacement of the string from its equilibrium at position x and time t.
(2 marks)
(b)
the speed of the wave
(1 mark)
(c)
ω is the angular frequency, equal to
2π
2π
, and k is the angular wave number, equal to
.
λ
T
(2 marks)
(d)
The wave is travelling in the –x direction because of the plus sign inside the cosine function.
(1 mark)
(e)
f ( x, t ) = A cos( k x + ω t )
Differentiate f ( x, t ) twice with respect to x as follows:
∂f
= − Ak sin ( k x + ω t )
∂x
∂2 f
= Ak 2 cos ( k x + ω t )
2
∂x
And, also twice with respect to t as follows
∂f
= − Aω sin ( k x + ω t )
∂t
∂2 f
= Aω 2 cos ( k x + ω t )
2
∂x
and substitute both into the wave equation which is
∂2 f
1 ∂2 f
=
∂x 2 v 2 ∂t 2
We get
1
Ak 2 cos(k x − ω t ) = 2 Aω 2 cos( k x − ω t )
v
The left and right hand side are equal if v =
ω
k
.
(3 marks)
(f)
This is the equation of a standing wave.
(1 mark)