Topic 3 – Simple Harmonic Motion

Key Words

Simple Harmonic Motion (SHM), equilibrium position, restoring force, displacement, velocity, acceleration
You are expected to know:
  • the characteristic features of simple harmonic motion ;
  • exchange of potential and kinetic energy;
  • the use of a = -(2pf)2x, x = Acos2pft, v = ±2pfΦ(A2 – x2);
  • Graphical representations linking displacement, velocity, acceleration, time, energy;

  • Graphical representations linking x, v, a and t .

  • Velocity as gradient of displacement-time graph.
    •

  There are four different kinds of motion that we can encounter in Physics:

Anything that swings or bounces or vibrates in a regular to-and-fro motion is said to oscillate. Examples include a swinging pendulum or a spring bouncing up and down.  It is said that the regularity of a swinging object was first described by a teenage Gallileo who watched a chandelier swinging during a church service in Pisa.

Simple Harmonic Motion (SHM) describes the way that oscillating objects move.  Consider a spring with a mass going from side to side.  A mass is mounted on a small railway truck, which is free to move from side to side, and there is negligible friction in the truck.

The rest or equilibrium position at O is where the spring would hold the mass when it is not bouncing.  A is the position where the spring is stretched the most, and B is where the spring is squashed most.

Question 1

Which way is the restoring force?  Why is there acceleration?  In which direction is the acceleration? 

ANSWER

At both A and B, the potential energy is at a maximum; the kinetic energy is zero.  

Question 2

Write down the formulae that describe kinetic energy and the elastic potential energy in a spring.  (The latter formula is NOT Ep = mgDh).

 ANSWER

As the mass passes the equilibrium position, there is zero potential energy, but maximum kinetic energy because this is the point at which the object has its greatest velocity (upwards or downwards).  Therefore there is an interchange between potential and kinetic energy.  The process is never 100 % efficient; some energy is lost as heat and the process is not indefinite.  

Question 3

How is the statement above consistent with the Law of Conservation of Energy?

 ANSWER 

We can write down a relationship between the acceleration, a, and the displacement, x.

F = ma and F = kx

Therefore

a = F/m = kx/m

So we are saying that the acceleration is proportional to the displacement from the equilibrium position.  However that is not the whole story.  Acceleration is a vector, so we must be careful of the direction.  The acceleration is towards the equilibrium position.  

Question 4

What do you think would happen if the direction of the acceleration were away from the rest position? 

ANSWER

For all cases:

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 harmonic.

In code we can write:

            a ΅ -x

ή a = - Kx where K is a constant.

The minus sign is important as it tells us that the acceleration is towards the equilibrium position. 

Some useful relationships for SHM

These relationships are derived by linking SHM to circular motion.  CLICK HERE if you want an explanation of why this is so.  You do not need to know this for the AQA exam (although other boards might ask you something on this).

Generally we measure the period, which is the time taken to make a complete oscillation or cycle.  The frequency is the reciprocal of the period:

f  = 1/T

Acceleration can be linked to displacement by:

a = - (2pf )2 x

This satisfies the condition for SHM that a = -Kx; in this case K = (2pf )2. A useful little dodge here is that p2 » 10.

Angular velocity is a quantity that is borrowed from circular motion.  It is the angle turned per second.  In SHM terms, we can consider it as the fraction of a cycle per second.  It can be, of course, greater than 1:

w = 2pf

In some texts you may see the equation for acceleration in SHM written as:

a = - w2 x

The speed at any point in the oscillation given by:

v2 = (2pf )2(A2 – x2)

ή v2 = 4p2f 2(A2 – x2)

ή v = 2pf Φ(A2 – x2)

In this relationship, A is the amplitude and s is the displacement from the equilibrium position.  If x = 0, v has a maximum value; if x = A, v = 0.  The velocity is 0 at each extreme of the oscillation.  So we can rewrite the equation as:

vmax = 2pfA

Note that the relationship only gives the speed (the magnitude of the velocity).  This is because the displacement is squared, so the minus sign disappears.  The relationship that gives velocity is:

v = Aw sin (wt)

 

The displacement, s, is given by:  

s =  ± A cos 2pft

The displacement can be shown graphically:

The plus and minus sign here tells us that the motion is forwards and backwards.  Which sign we give for direction is up to the individual.  Generally left to right is forwards.

All these equations are true for any simple harmonic motion.  We can show the relationships graphically by showing displacement, velocity, and acceleration against time:

These graphs are sinusoidal.  The displacement is p/2 radians (90 o or Ό cycle) behind the velocity.   The displacement and acceleration are p radians out of phase.

 Worked Example

A particle moving with simple harmonic motion has velocities 4 cm/s and 3 cm/s at distances of 3 cm and 4 cm respectively from the equilibrium position.  What is the amplitude of the oscillation?  What is the velocity of the particle as it passes the equilibrium position?

We know that v2 = 4p2f 2(A2 – x2) ή v = 2pf Φ(A2 – x2)  [A - amplitude, s – displacement]

When x = +3 cm, v = 4 cm/s; when x = + 4 cm, v = 3 cm/s.  We don’t know what f is.

We can substitute the numbers into the equations:

                                                42 = 4p2f 2(A2 – 32) ……[1]

                                                            32 = 4p2f 2(A2 – 42) ……[2]

To get rid of the 4p2f 2 we need to divide [1] by [2]:

                        16 = A2 – 9

                          9    A2 – 16

 Rearranging: ή 16(A2 – 16) = 9(A2 – 9)

                        ή 16A2 – 9A2  = 256 – 81

                        ή        7A2 = 175

                        ή          A2 = 175 Έ 7 = 25

                        ή          A = 5 cm

Now we can find the period by finding w.  Since w = 2pf, we can rewrite the equation v2 = 4p2f 2(A2 – x2)  as v2 = 4w2(A2 – x2) :

                        16 = w2 (25 – 9) = 1 rad/s

 ή T = 2p Έ 1 = 6.28 s

Now we can work out the velocity at the equilibrium point (s = 0).

                        v2 = 1(25 – 0) = 25

ή v = 5 cm/s

 

Question 5

A pendulum has a period of 3.0 s and an amplitude of 0.1 m.  What is its frequency and what is its maximum acceleration? 

ANSWER
Question 6

A punch-bag of mass 0.60 kg is struck so that it oscillates with SHM.  The oscillation has a frequency of 2.6 Hz and an amplitude of 0.45 m.  What is:

(a)    the maximum velocity of the bag;

(b)   the maximum kinetic energy of the bag?

(c)    What happens to the energy as the oscillations die away?

ANSWER

 

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