A LEVEL

17 Oscillations
Simple harmonic oscillations
Terminology
Consider a ruler clamped to a bench, pulled downwards and released so that it
vibrates; a pendulum swinging backwards and forwards; a mass on the end of a spring
bouncing up and down. These are all examples of oscillating systems (Figure 17.1).

(a) (b) (c)

▲ Figure 17.1 Oscillating systems

One complete oscillation is when an object moves:

» from its equilibrium position to its maximum displacement in one direction
» back through the equilibrium position to the maximum displacement in the
opposite direction
» and back once more to the equilibrium position
This is shown in Figure 17.1(c).
» The period, T, is the time taken for one complete oscillation of an object.
» The frequency, f, is the number of oscillations per unit time.
» The displacement, x, is the vector distance from the equilibrium position at an
instant.
» The amplitude, x0, is equal to the magnitude of the maximum displacement of an
object from its mean position.
» The angular frequency, ω, is equal to 2πf.
» The phase difference is the fraction of a cycle between two oscillating objects,
expressed in either degrees or radians. (See p. 57.)
It is worth remembering the following relationships, which you may recognise from
the work on circular motion:
1 2π
F= ω = 2πf ω=
T T

Simple harmonic oscillations

In the examples above, the objects vibrate in a particular way known as simple
harmonic motion (s.h.m.). There are many other types of oscillations. For instance,
a conducting sphere will oscillate between two charged conducting plates – but not
with simple harmonic motion.

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 17 Oscillations

The conditions required for simple harmonic motion are:

STUDY TIP
» the magnitude of the acceleration is proportional to the displacement from a In the conducting
fixed point sphere example, when
» the direction of the acceleration is always in the opposite direction to the the charged sphere
displacement bounces back and forth
» this means that the acceleration is always directed towards the fixed point between two parallel
charged conductors,
Simple harmonic motion can be investigated using a position sensor connected to a the magnitude of
datalogger (Figure 17.2). the electric force is
constant as it moves
between the plates.
The first condition
(acceleration ∝
Position sensor displacement) is not
Pendulum To datalogger followed. Therefore,
this cannot be simple
harmonic motion.
▲ Figure 17.2

The displacement against time graph can be deduced from the trace on the
datalogger (Figure 17.3).

▲ Figure 17.3

As with any displacement–time graph, the velocity is equal to the gradient of the
graph; the acceleration is equal to the gradient of the velocity–time graph. STUDY TIP
The displacement–time
Displacement graph can be started at
any point on the cycle.
Here, the equilibrium
Time position is chosen as
the starting point.
Velocity
Other books might
choose maximum
displacement, in which
Time case the displacement
curve would be a cosine
Acceleration
curve, the velocity
curve would be a minus
sine curve and the
Time acceleration would be a
minus cosine curve.
▲ Figure 17.4

Table 17.1 describes the displacement, the velocity and the acceleration at different
points during an oscillation, with reference to Figure 17.4.
» A more detailed analysis of the graphs shows that the velocity–time graph is
obtained by plotting a graph of the gradient of the distance–time graph against
time. The equation for this graph is v = v0 cos ωt
» The acceleration–time graph is obtained by plotting a graph of the gradient of
the velocity–time graph against time.
» Remember that the velocity is obtained from the gradient of the distance–time
graph and acceleration is from the gradient of the velocity–time graph. Refer
back to p. 19.

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 ▼ Table 17.1

Point in cycle Displacement Velocity Acceleration

t=0 Zero Maximum in one Zero
direction
¼ cycle on Maximum in one Zero Maximum in the
from t = 0 direction opposite direction
from the displacement
½ cycle on Zero Maximum in the Zero
from t = 0 opposite direction
from before
¾ cycle on Maximum in the Zero Maximum in the
from t = 0 opposite direction opposite direction
to before from the displacement
1 cycle on Zero Maximum in the Zero
from t = 0 original direction

Equations for simple harmonic motion

If you look at the graphs of simple harmonic motion (s.h.m.), you will see that they
are of the form of sine (or cosine) graphs. The conditions for s.h.m. give the
following proportionality:
a ∝ −x
where a is the acceleration and x is the displacement. STUDY TIP
The minus sign comes in because the acceleration is in the opposite direction from a = −x0ω2 sin ωt and
the displacement. x = x0 sin ωt
This leads to the equation: Dividing the first equation
by the second gives:
a = −ω2 x
a −x0ω2 sin ωt
where ω is the angular frequency. x = x0sin ωt
This equation describes simple harmonic motion. The graphs in Figure 17.4 are Cancelling the x0 and the
‘solutions’ to this equation. If you look at those graphs, you will see that they have sin ωt top and bottom
gives:
a sine (or cosine) shape. The precise equations that they represent are:
a
» displacement: x = x0 sin ωt = −ω2
x
» velocity: v = x0ω cos ωt
which gives:
» acceleration: a = −x0ω2 sin ωt
where x0 is the amplitude of the oscillation. a = −ω2 x

Look at the equations for displacement and acceleration. Can you see that they fit in
with the equation a = −ω2 x?
The velocity of the vibrating object at any point in the oscillation can be calculated
using the formula:
v = ± ω x02 − x2
It follows that when x = 0 (i.e. the displacement is zero) the velocity is a maximum
and:
v0 = ±ωx0

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 17 Oscillations

WORKED EXAMPLE
A mass on the end of a spring oscillates with a period of 1.6 s and an amplitude
of 2.4 cm. Calculate:
a the angular frequency of the oscillation
b the maximum speed of the mass
c the maximum acceleration
d the speed of the mass when its displacement from the equilibrium position
is 0.60 cm

Answer
1 1
a ƒ= = Hz
T 1.6
1
ω = 2πf = 2π × = 3.9 rad s−1
1.6
b vmax = ωx0 = 3.9 × 2.4 = 9.4 cm s−1
c a = −ω2 x
amax = ω2 x0 = 3.92 × 2.4 = 37 cm s−2

d v = ± ω √x02 − x2 = 3.9 × √2.42 − 0.62 = 9.1 cm s−1

NOW TEST YOURSELF

1 A stone of mass 0.80 kg attached to the bottom of a vertical spring
oscillates with a time period of 1.8 s and an amplitude of 4.4 cm. Calculate:
a the frequency of the stone
b the angular frequency of the stone
c the maximum acceleration of the stone
2 The period of a simple pendulum is given by the formula: T = 2π√l/g.
a Calculate the period of a pendulum of length 25 cm.
b Calculate the acceleration of the bob when the displacement is 3.0 cm.
3 A particle vibrates with simple harmonic motion of amplitude 5.0 cm and
frequency 0.75 Hz. Calculate the maximum speed of the particle and its
speed when it is 2.5 cm from the central position.

s.h.m. and circular motion

The introduction of ω should have reminded you of circular motion. The description
of the following experiment shows the relationship between circular motion and
simple harmonic motion (Figure 17.5).

Screen

Pendulum

Rod

Turntable

Amplitude

▲ Figure 17.5

» A rod is set up on a turntable, which rotates.

» A pendulum is set swinging with an amplitude equal to the radius of the rotation
of the rod.
» The speed of rotation of the turntable is adjusted until the time for one
revolution of the turntable is exactly equal to the period of the pendulum.
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 » The whole apparatus is illuminated from the front so that a shadow image is
formed on a screen.
» It is observed that the shadow of the pendulum bob moves exactly as the shadow
of the rod.
» This shows that the swinging of the pendulum is the same as the projection of
the rod on the diameter of the circle about which it rotates (Figure 17.6).

ωt
ωt

x
x0

▲ Figure 17.6

You should now understand the close mathematical relationship between circular
motion and simple harmonic motion.

Energy in simple harmonic motion

Kinetic energy and potential energy
During simple harmonic motion, energy is transferred continuously between kinetic
and potential energy:
» In the case of a pendulum, the transfer is between kinetic and gravitational
potential energy.
» In the case of a mass tethered between two horizontal springs, the transfer is
between kinetic and strain potential energy.
(a) (b) (c)
Ek Ep Total energy

–x0 x0 –x0 x0 –x0 x0

▲ Figure 17.7 (a) The variation of kinetic energy with displacement, (b) the variation of
potential energy with displacement, (c) the total energy with displacement
» The speed of the particle is at a maximum when the displacement is zero so that
the kinetic energy is maximum at this point and the potential energy is zero
(Figure 17.7a,b).
» At maximum displacement, the speed, and hence the kinetic energy, is zero and
the potential energy is maximum (Figure 17.7a,b).
» The important point is that in any perfect simple harmonic oscillator, the total
energy is constant. This means that the sum of the kinetic and potential energies
remains constant throughout each oscillation (Figure 17.7c).
The equations that link the kinetic energy and the potential energy to the
displacement are:
» kinetic energy: Ek = ½mω2(x02 − x2)
» potential energy: Ep = ½mω2 x2
» total energy at any point in the oscillation: Ek + Ep ⇒ E = ½mω2 x02
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 17 Oscillations

WORKED EXAMPLE
2π
A clock pendulum has a period of 2.0 s and a mass T = 2.0, therefore ω = =π
2
of 600 g. The amplitude of the oscillation is 5.2 cm.
Calculate the maximum kinetic energy of the Ek = ½mω2 x02 = 0.5 × 0.600 × π2 × (5.2 × 10 −2)2
pendulum and, hence, its speed when it is travelling Ek = 8.0 × 10−3 J
through the centre point.
Ek = ½mv2
Answer 2Ek 2 × (8.0 × 10−3)
v= m = = 0.16 m s−1
Ek = ½mω2(x02 − x2); for maximum speed the 0.6
displacement = 0

NOW TEST YOURSELF

4 A pendulum has a length of 5.0 m and an amplitude of 12 cm. The bob has a
mass of 0.50 kg. Calculate:
a the maximum speed of the pendulum bob
b the maximum restoring force on the bob
c the maximum kinetic energy of the bob
d the total energy of the system

Damped and forced oscillations, resonance

Damping
» Up to this point, we have only looked at perfect simple harmonic motion, where
the total energy is constant and no energy is lost to the surroundings.
» In this situation, where the only force acting on the oscillator is the restoring
force, the system is said to be in free oscillation.
» In real systems, some energy is lost to the surroundings for a variety of reasons,
including due to friction and/or air resistance.
» This always acts in the opposite direction to the restoring force.
» The result is that the amplitude of the oscillations gradually decreases. This is
called damping (Figure 17.8).

Displacement

0
Time

▲ Figure 17.8 A lightly damped oscillation

» The decay of the oscillation follows the exponential decay (see p. 161).
» The period, however, remains constant until the oscillation dies away
completely.
» Figure 17.8 shows light damping – the oscillation gradually fades away.

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 » If the damping is increased, we eventually reach a situation where no complete
oscillations occur and the displacement falls to zero. When this occurs in the Heavy damping

Displacement
minimum time, the damping is said to be critical (see Figure 17.9).
» More damping than this is described as heavy damping and the displacement
0
only slowly returns to zero (see Figure 17.9). Time
Critical
damping
Examples of damped oscillations
▲ Figure 17.9

Chassis

Shock absorber

Spring

Hub of wheel

▲ Figure 17.10 The suspension on a car relies on critically damped harmonic motion

A car suspension (Figure 17.10) operates in a critical damping mode in order to bring
the displacement back to zero in the shortest possible time without oscillations.
A heavily damped suspension leads to a hard ride, with energy given to the car by
bumps not being absorbed as efficiently.

Forced oscillations
» In Chapter 8, you met the idea of stationary waves formed on a string
(pp. 67–68). This is an example of a forced oscillation.
» An extra periodic force is applied to the system. This periodic force continuously
feeds energy into the system to keep the vibration going.
» You will have observed how the amplitude of the vibrations of the waves on a
string changes as the frequency of the vibrator is changed:
» a small amplitude at very low frequencies
» gradually increasing to a maximum as the frequency is increased
» then reducing again as the frequency is increased further (Figure 17.11)

Amplitude

Frequency
Resonant frequency
KEY TERMS
▲ Figure 17.11 The amplitude of a forced oscillation at different frequencies The natural frequency
» This is an example of resonance. of a vibration is the
frequency at which an
» When the driving frequency is the same as the natural frequency of oscillation object will vibrate when
of the string, then it gives the string a little kick at the right time each cycle allowed to do so freely.
and the amplitude builds up.

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 17 Oscillations

Resonance can be demonstrated using Barton’s pendulums (Figure 17.12).

Metre rule Flexible wire

Paper cone
Driving
Curtain ring pendulum
Strong cotton

▲ Figure 17.12 Barton’s pendulums

REVISION
» The driving pendulum causes the paper-cone pendulums to vibrate.
» Only the pendulums of a similar length to the driving pendulum show any ACTIVITY
significant oscillation.
» All the pendulums vibrate with the same frequency, which is the frequency of the Look for repetitive
vibrating systems,
driving pendulum (not their own natural frequencies).
for example, a ball
» This is a general rule for all forced oscillations.
bouncing.
Decide for yourself
NOW TEST YOURSELF and justify whether
or not the system
5 Explain the difference between critical damping and heavy damping. vibrates with simple
6 Explain what is meant by the term resonance. harmonic motion.

END OF CHAPTER CHECK

In this chapter, you have learnt to: » analyse and interpret graphical
» understand and use the terms displacement, representations of the variations of
amplitude, period, frequency and phase displacement, velocity and acceleration for
difference in the context of oscillations simple harmonic motion
» express the period in terms of frequency » describe the interchange between kinetic
» express the frequency in terms of angular and potential energy of a system undergoing
frequency simple harmonic motion
» understand that simple harmonic motion » understand that a resistive force acting
occurs when acceleration is proportional to on an oscillating system causes damping
displacement from a fixed point and in the » understand and use the terms light, critical
opposite direction and heavy damping and sketch displacement–
» use the formula a = −ω2 x time graphs illustrating these types of
» recall and use x = x0 sin ωt as a solution to damping
the above equation » understand that resonance involves a
» use the equations v = v 0 cos ωt and maximum amplitude of oscillations and that
v = ± ω √x02 − x2 this occurs when an oscillating system is
forced to oscillate at its natural frequency

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