AL Physics/S.H.M./P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
Form Six AL PHYSICS
Simple Harmonic Motion
Simple Harmonic
Motion
Isochronous oscillation. Acceleration a = -ω2x,
displacement x = A sin ωt (or A cos ωt). Simple
harmonic motion developed through analysis of
uniform motion in a circle (rotating vector
model). Applications to include the simple
pendulum and loaded spring. Quantitative
treatment of kinetic and potential energy. Phase
lead and phase lag through rotating vector model.
E2. Study of the motion of one
simple harmonic oscillator
(e.g. ticker tape attached to
long pendulum).
A. Spring-mass system
1. Hooke’s Law
- Consider a mass m attached to a spring
with natural length l0. Let the mass be
l
0
x =0
m
displaced a distance x from its original
x
position at x = 0, (i.e. the spring extended
from l0 to l0 + x). As the agent exerts a
force Fext on the particle, the spring exerts
F
s
m
an opposing force Fs. It is found that this force Fs is directly proportional to the
extension x of the spring, i.e. F ∝ x. Furthermore, Fs always acts in the opposite
direction to x, so
Fs ∝ - x
= - kx
(Hooke’s Law)
where k is known as the force constant of the spring.
- The constant k is a measure of the force necessary to produce a given stretching of
the spring; stiffer springs have larger values of k.
- The minus sign indicates the fact that the direction of the force Fs is always opposite
to the direction of the displacement of the particle. When the spring is stretched, x
> 0 and Fs is negative, when the spring is compressed, x < 0 and Fs is positive.
(Hence, the force exerted by the spring is a restoring force; it always tends to restore
the particle to its original position.)
F
ext
AL Physics/S.H.M./P.2
2. Potential Energy Stored in a stretched spring
x =0
- Now we are interested in calculating the work
m
done by the external force, which is stored in
the stretched spring.
x
- If the mass m is displaced a distance x from its
original position at x = 0 slowly (i.e. the
F
s
F
m
ext
external force is always approximately equal to
the spring force, so that the mass is nearly in equilibrium at all times).
-
W.D. =
∫
=
∫
x
0
x
0
Fext dx
kx dx
Fext
[Fext = -Fs = -(-kx) = kx]
= ½kx2
kx
- In a mass-spring system, whenever the mass is
displaced a distance x from its reference position (i.e.
the unstretched position), the potential energy of the
system is ½kx2. The same result is obtained whether x
∆x
is +ve or -ve, i.e. whether the spring is stretched or
compressed by a given amount x, the stored energy is
the same.
3. Springs in Series and Parallel
a. In series
Let k be the equivalent force constant of the spring system.
T = k(e1 + e2)
T = k1e1
T = k2e2
T T
hence, T = k ( + )
k1 k2
⇒
1 1
1
= +
k k1 k2
If
k1 = k2,
k = k1/2
k1
k2
T
(the force constant of the system is smaller !!)
T
x
AL Physics/S.H.M./P.3
b. In parallel
Let k be the equivalent force constant of the spring system.
F1 = k1e
F2 = k2e
⇒ F = F1 + F2
k1
F1
F
= (k1 + k2)e
Hence, k = k1 + k2
k2
if k1 = k2
F2
k = 2k1
B. Mass-Spring Oscillation
-
Consider a mass m attached to a spring with force constant k. The mass is displaced
to x and released with zero velocity at time t = 0. Hence the mass is under the
influence of the force of the spring Fs. And the
Equilibrium position
x =0
x
equation of motion of the mass is given by
ma = Fs
k
= - kx
⇒
-
a= −
m
k
x
m
The subsequent time variation of the displacement x, velocity v, and acceleration a are
as shown below:
2π
x = A0 cos( t ) = A0 cos ωt
T
x
A0
t
0
T
A0
2π
v = − v0 sin( t ) = - v0 sin ωt
T
v
v0
0
T
2T
t
v0
a
a = − a0 cos(
2π
t ) = -a0 cos ωt
T
T : period of the oscillation
A0 : Amplitude
a0
0
a0
T
t
AL Physics/S.H.M./P.4
-
To verify x = A0 cos(
2π
t ) (or x = A0 cos ωt) is a solution of the equation of motion,
T
differentiate x with respect to time t.
dx
= v = −ωA0 sin(ωt )
dt
[v0 = ωA0]
d2x
= a = −ω 2 A0 cos(ωt )
2
dt
[a0 = ω2A0]
= - ω2x
⇒
ω2 = k/m, ω = √(k/m)
or
2π/T = √(k/m)
hence T = 2π
m
k
Note that the period is independent of the amplitude A0. The motion is said to be
isochronous.
-
S.H.M.: If the acceleration of a body is directly proportional to its distance from a
fixed point and is always directed towards that point, the motion is simple harmonics.
a ∝ -x
-
Remark:
In general, the solution of the differential equation
d2x
= −ω 2 x
2
dt
x = A0 cos (ωt + φ)
is
where
v=
dx
= −ωA0 sin(ωt + φ)
dt
a=
d2x
= −ω 2 A0 cos(ωt + φ)
2
dt
A0 : Amplitude
φ : initial phase
The values of A0 and φ are determined by the initial condition of the oscillation. For
example, if the mass is initially at rest at the equilibrium position and the mass are set
into oscillation by a sudden “kick”, at time t = 0, x = 0, v = -v0 and a = 0. In order to
satisfy these initial condition, φ should be
Equilibrium position
x =0
+ve x
equal to π/2 and the time variation of x, v and
a are then
x = A0 cos(ωt +π/2) = -A0 sin ωt
v = −ωA0 sin(ωt + π / 2) = -ωA0 cos ωt
k
"kick"
m
AL Physics/S.H.M./P.5
a = −ω 2 A0 cos(ωt + π / 2) = ω2A0 sin ωt
(For the determination of the amplitude A0, it will be discussed later.)
C. Examples
1. A cylindrical rod of mass m floats vertically in equilibrium in a liquid of density ρ.
The cross-sectional area of the rod is A. If the rod is pushed down and release, prove
that the subsequent motion is a S.H.M.
net force acting on the rod = upthrust = xAρg (upward)
hence,
ma = - xAρg
a=−
Aρg
x
m
a ∝ -x
i.e.
x
B
(S.H.M. !!)
Aρg
m
ω=
B : equilibrium position
m
T = 2π/ω = 2π
Aρg
2. Vertical Spring
- Consider a spring with a mass m hung at its end vertically.
- Choose the origin x = 0 corresponding to the equilibrium position, i.e. the position
where ke - mg = 0.
- With the mass pull down slightly (x), the total extension = x + e
Hence, the net force (upwards) acting on the mass
= -k(x + e) + mg
= -kx - ke + mg
(ke = mg)
= -kx
natural
2
Hence,
m
length
d x
= − kx
dt 2
d2x
k
=− x
2
m
dt
(S.H.M. !!)
e
x =0
x
ω=
⇒
k
m
T = 2π
m
k
(same as that in the horizontal spring system)
AL Physics/S.H.M./P.6
3. Simple pendulum
d 2θ
= − mg sin θ
dt 2
ml
for small θ, sin θ ~ θ
θ
l
d θ
= − gθ
dt 2
2
l
d 2θ
g
θ
2 = −
dt
l
ω=
g
l
⇒
(S.H.M. !!)
mg
T=
2π
l
= 2π
g
ω
Remark: The motion of the bob is S.H.M. only if the oscillations are of small
amplitude (i.e. small θ, usually θ should not exceed 10°)
4. Oscillations of a liquid in a U-tube
- The liquid (total length 2h) on one side of a U-tube is depressed by blowing gently
down that sides, the levels of the liquid will oscillate for a short time about their
respective initial positions O & C before finally coming to rest.
-
net force = -2xAρg
hence
⇒
d2x
m 2 = −2 xAρg
dt
2hAρ
d x
= −2 xAρg
dt 2
d2x
g
=− x
2
h
dt
ω=
x =0
2
g
h
(S.H.M. !!)
⇒
T = 2π
h
g
x
O
C
AL Physics/S.H.M./P.7
D. Simple Harmonic Motion and Uniform Circular Motion
-
Consider a point mass m undergo a uniform circular motion with angular velocity ω.
-
The x-projection of the displacement x, velocity vx and acceleration ax are given as
x = A cos (ωt + φ)
ω
y
v = - ωA sin (ωt + φ)
v
a = - ω2A cos (ωt + φ)
hence,
t
A
a = - ω2x
a ∝ -x
ωt
t =0
φ
(S.H.M. !!)
x
A
φ : initial phase of
the motion
-
A S.H.M. can be described as the projection of uniform circular motion along a
diameter of the circle.
Example 1:A mass which is attached to a spring is displaced 0.160 m from its equilibrium
position and released with zero initial speed. After 1.10 s its displacement is found to
be 0.1 m on the opposite side, and it has passed the equilibrium once during this
interval. Find (a) the amplitude; (b) the period; (c) frequency.
Solution:(a) Amplitude = 0.160 m
(b) φ = π - cos-1(0.1/0.16) = 2.25 rad
φ
hence, ω = φ/1.1 = 2.04 rad/s
T = 2π/ω = 2π/2.04 = 3.08 s
0.1 m
0.160 m
(c) f = 1/T = 0.325 Hz
Example 2:A body is vibrating with simple harmonic motion of amplitude 15 cm and frequency 4
Hz. Compute
(a) the maximum values of the acceleration and velocity,
(b) the acceleration and velocity when the coordinate is 9 cm, and
(c) the time required to move from the equilibrium position to a point 12 cm distant
from it.
AL Physics/S.H.M./P.8
Solution:(a) amax = ω2A
= (8π)2 × 0.15
θ ω = 4 × 2π
= 8π rad/s
ωA
= 94.7 ms-2
15 cm
vmax = ωA
= 8π × 0.15
12 cm
= 3.77 ms-1
15 cm
β
ω2A
θ
152 − 9 2
= 12 cm
9 cm
15 cm
(b)
a (x = 9 cm) = ω2A cos θ
9
15
= 94.7 ×
= 56.85 ms-2
(direction : to the left)
v (x = 9 cm) = ωA sin θ
= 377
. ×
12
15
= 3.02 ms-1
(c) β = sin −1
12
15
= 53.13º
∴
5313
. !
t
=
T
360!
t =
5313
. ! 1
×
360 ! 4
= 0.037 s
Example 3:A body of mass 10 g moves with simple harmonic motion of amplitude 24 cm and
period 4 s. The coordinate is +24 cm when t = 0. Compute
(a) the position of the body when t = 0.5 s,
(b) the magnitude and direction of the force acting on the body when t = 0.5 s,
(c) the minimum time required for the body to move from its initial position to the
point where x = -12 cm.
(d) the velocity of the body when x = -12 cm.
AL Physics/S.H.M./P.9
Solution:(a)
t
θ
! =
360
T
T=4s
0.5
θ =
× 360 !
4
t = 0.5 s
2
mω A
= 45º
θ
t=0
x = A cos θ
= 24 × cos 45º
24 cm
= 17 cm
(b) force = mω2A cos θ
= 10 × 10 − 3 × (
2π 2
) × 0.24 cos 45!
4
= 4.2 × 10-3 N
(c) θ = 90! + sin −1
(direction : to the left)
12
24
= 120º
t
ωA
!
120
t
=
T
360!
β
β
1
t = × 4 = 1.33 s
3
θ
12 cm
t=0
24 cm
(d) v (x = -12 cm) = ωA cos β
=(
2π
)24 cos 30 !
4
= 32.6 cm/s
= 0.33 m/s
Example 4:A small block is executing simple harmonic motion in a horizontal plane with an
amplitude of 10 cm. At a point 6 cm away from equilibrium, the velocity is 24 cm/s.
(a) What is the period?
(b) What is the displacement when the velocity is ±12 cm/s?
(c) If a small object placed on the oscillating block is just on the verge of slipping at
the endpoint of the path, what is the coefficient of friction?
AL Physics/S.H.M./P.10
Solution:ωA sin θ = 24
(a)
(
2π
8
) × 10 ×
= 24
T
10
θ
ωA
β
10 8
T = 2π ×
×
24 10
=
2
π
3
24 cms-1
ωA
β
6 cm
12 cms-1
= 2.09 s
10 cm
ωA sin β = 12
(b)
(
2π
) × 10 sin β = 12
2.09
β = 23.6º
x = A cos 23.6º
= 10 cos 23.6º
= 9.2 cm
(c) at the endpoint of the path:
a = ω2A
=(
2π 2
) × 01
.
2.09
= 0.9 ms-2
the acceleration is provided by the frictional force
i.e.
mgµ = ma
⇒
µ =
=
a
g
0.9
9.8
= 0.09
θ
102 − 62
= 8 cm
AL Physics/S.H.M./P.11
E. Energy of S.H.M.
-
Consider the oscillation of a spring-mass system.
-
In the oscillation, there is a constant interchange of energy
A
between K.E. and P.E.
-
ωt + φ
x
Variation of K.E. and P.E. with displacement
P.E. = ½kx2
K.E. = ½mv2
= ½mω2A2sin2(ωt + φ)
=
Energy
1
( A2 − x 2 )
mω 2 A2
2
A2
1_
kA 2
2
= ½mω2(A2 - x2)
2
2
= ½k(A - x )
P.E.
[ω = √(k/m)]
K.E.
Total energy = P.E. + K.E.
= ½kx2 + ½k(A2 - x2)
= ½kA2
-
A
A
displacement x
(constant with displacement)
Variation of K.E. and P.E. with time
P.E. = ½kx2 = ½kA2cos2ωt
Energy
1_
kA 2
2
K.E.
K.E. = ½mv2 = ½mω2A2sin2ωt
= ½kA2sin2ωt
P.E.
Total energy = P.E. + K.E.
= ½kA2
(constant with time)
0
T/2
T
Example:- A particle moving with s.h.m. has velocities of 4 cm/s and 3 cm/s at
distances of 3 cm and 4 cm respectively from its equilibrium position.
find (a) the amplitude of the oscillation, (b) the period, (c) the velocity of
the particle as it passes through the equilibrium position.
F. Concept of Phase
1. Consider two sinusoidal functions
ψ1 = A1 cos (ωt + φ1)
ψ2 = A2 cos (ωt + φ2)
phase difference between ψ1 and ψ2 is given by (φ1 - φ2).
time
AL Physics/S.H.M./P.12
a. In phase (i.e. φ1 - φ2 = 0)
ψ1
- If φ1 - φ2 = 0, the two function ψ1 and ψ2 are said to be
ψ2
in phase.
- At the same instant, ψ1 and ψ2 pass through the zero
t
0
position and reach the maximum (minimum) values.
b. Anti-phase (180° out of phase, i.e. φ1 - φ2 = 180°)
ψ1
- If φ1 - φ2 = 180°, ψ1 and ψ2 are said to be anti-phase or
180° out of phase.
- At the same instant, ψ1 and ψ2 pass the zero position.
t
0
However, whenever ψ1 is maximum, ψ2 is minimum,
ψ2
and vice versa.
c. φ1 - φ2 ≠ 0 and ≠ 180°
- ψ1 leads ψ2 by (φ1 - φ2) or ψ2 lags ψ1 by (φ1 - φ2)
- Example:
ψ1 = A1 cos (ωt - 30°)
ψ2 = A2 cos (ωt + 15°)
φ1 - φ2 = -30° - 15° = -45°
Hence, ψ1 lags ψ2 by 45° or ψ2 leads ψ1 by 45°
- The two functions no longer pass through the zero point at the same time. The
phase difference φ1 - φ2 indicates the difference between the time at which the two
functions pass through the zero point (or reach the maximum value).
- Rotating vector model
A sinusoidal function ψ = A cos (ωt + φ) can be viewed as the x (or y) projection
of an arrow of length A rotating at an angular velocity ω. And the initial (i.e. at
time t = 0) angular position of this arrow is φ. The phase difference between any
two sinusoidal functions (same angular velocity) is just the angle between the two
rotating arrows.
ω
ψ1
ψ2
θ
t
ψ1
ψ2
ψ2 leads ψ1 by θ (or ψ1 lags ψ2 by θ)
** Here the functions are viewed as the “y-projection of the rotating vector”.
- Example:-
AL Physics/S.H.M./P.13
ω
ψ1
ψ2
5
3
θ
2.5
5
-1
ψ1
3
ψ2
t
Phase difference between ψ1 and ψ2 = sin-1(2.5/3) + sin-1(1/5) = 68° (or 1.2 rad)
Hence ψ2 leads ψ1 by 1.2 rad (or ψ1 lags ψ2 by 1.2 rad)
2. In Simple Harmonic Motion
ω
x = A0 cos (ωt + φ)
v = -ωA0 sin (ωt + φ)
= ωA0 cos (ωt + φ + π/2)
ωA0
A0
v
x
a = -ω A0 cos (ωt + φ)
2
= ω2A0 cos (ωt + φ + π)
a
The acceleration leads the velocity by π/2.
The velocity leads the displacement by π/2.
ω2 A 0