Further Mechanics Tutorial 7 - Simple Harmonic Systems


1.  Mass on a spring

The extension of a spring is directly proportional to the force (Hookes Law).  Consider a mass, m,  put onto a spring of spring constant k so that so that it stretches by an extension l.

The force on the spring = mg, and the stretching tension = kl.

                        mg = kl

Question 1

Why does the spring stretch when you add a mass to it?


Suppose the spring is pulled down by a distance x below the rest position.  Now the stretching force become k(l + x).  This is also the tension in the spring acting upwards.  So the restoring force, Fup = k(l + x) mg This is because mg is the weight, which always acts downwards.

  Since kl = mg, we can write:

  Fup = kl + kx kl = kx

  We can now apply Newton II to write:

  -kx = ma  (The negative sign tells us that the force is upwards)

  We know from SHM that a = - (2pf )2 x, so we can write:

a = -kx/m = - (2pf )2 x

  So we can rearrange this to say that:

 (2pf )2 = k/m

Since a = - (2pf )2 x we can say that the condition for SHM is satisfied in this system, as long as Hookes Law is obeyed.  Since T = 1/f, we can now write down an expression to relate the period with the mass and the spring constant.

T = 2p(m/k)

This tells us that if we want to double the period, the mass has to be increased by four times. 

If we plot a graph of T2 against m we will get a straight line, since T2 = 4p2 (m/k).  The gradient will be 4p2/k which we can approximate to 40/k, since p2 10.

The relationship of the graph suggests that the line should cut through the origin.  However we find that it does not.  This is due to the mass of the spring itself; the effective mass of the spring is about 1/3 the actual mass of the spring itself.  However if the mass on the spring is very much bigger than the mass of the spring, this effect is negligible.

Worked Example

A light spiral spring is loaded with a mass of 50 g and extends by 10 cm.  What is the period of small vertical oscillations if the acceleration due to gravity is 10 m/s2?

We need to work out the spring constant using Hookes Law F = ke

k = F/e = 0.05 kg 10 m/s2 = 5.0 N/m

                    0.1  m

Now we can use T = 2p(m/k) to work out the period:

  T =  2p(0.05 kg) = 2p(0.01) = 0.2 p = 0.63 s

               5 N/m

A common bear trap is to forget to take the square root.

Question 2

A spring has a spring constant of 80 N/m.  A mass of 0.5 kg is placed on the spring and the spring is allowed to oscillate.  What is the frequency of the oscillation?



2. The Simple Pendulum

Consider a small bob of mass m hanging from a very light string, length l , which in turn hangs from a fixed point.  If it is pulled to one side through a small angle q, it will swing with a to-and-fro movement in the arc of a circle.


As weight mg is a vector, we can break it into its two components, mg cos q and mg sin q.

At point A the bob accelerates with an acceleration a due to the force mg sin q.  We can apply Newton II to write:

-mg sin q = ma             [negative sign as the force is directed to equilibrium position]

If q is small and in radians, we can say that sin q = q.  This does not work for degrees. 

Make sure your calculator is set to radians.

We can measure the chord OA, which is the displacement.  Remember that displacement is the straight-line distance between two points.  Now we can say that the displacement, s = l sin q.  So we can therefore rewrite this as s = lq.  So now we can write:

 -mgq = -mg(s/l) =  ma             [m terms cancel out]

a = -gs/l = -(2pf )2s

The relationship a = -(2pf )2s arises because this is a simple harmonic oscillator. 

In this case (2pf )2 = g/l because the s terms cancel out.

We know that period is given by T = 1/f.

         T = 1 = _2p_  = 2p(l/g)

   f    (g/l)

So we can write the formula linking period with length and gravity constant as:

T = 2p(l/g)

Notice that T is independent of amplitude or mass of the bob.  If a pendulum clock were taken to the moon, its time-keeping would be somewhat altered.

Question 3

How much would the time keeping of a pendulum clock be affected by taking it to the moon?  Gravity on the moon is 1.6 N/kg, compared with 10 N/kg on earth.

Question 4

Would a bouncing spring oscillator be affected by the weakened gravity field on the Moon?


If we plot a graph of T2 against l, we get a straight line of which the gradient is 4p2/g (the value of which would be approximately 4).  To measure g we divide 4p2 by the gradient we get.  We can get a relatively accurate determination of this if we:


Worked Example

A simple pendulum has a period of 2.0 s and amplitude of swing of 5.0 cm.  What is the maximum velocity of the bob?  What is the maximum acceleration?
Velocity is at a maximum when the equilibrium position is reached.

v2 = w2(A2 x2    x = 0, A = 5 cm.

We need to know w. 

w = 2p/T = 2p rad 2.0 s = p rad/s

Now we can work out the velocity:

v2 = p2(52 02) = p2 25 = 247 cm2 s-2

v = 15.7 cm/s

We can use a = -w2s.  Acceleration is at a maximum when s = A = 5.0 cm

a = -p2 5.0 = - 50 cm/s2.  The negative sign tells us that the direction of the acceleration is to the rest position.


Question 5

What is the time period for a pendulum of length 4.6 m.  Take g = 9.8 m/s2

Question 6

The amplitude of the swing of the pendulum in question 11 is 0.50 m.  What is the maximum acceleration?  What is the maximum velocity? 



Energy in Simple Harmonic Motion

There is constant interchange between kinetic and potential energy as the pendulum (or other oscillator) swings to-and-fro.  If the system does not have to work against restrictive forces, such as friction, the total energy will remain constant.

This is the most likely level you need.  Click HERE if you want to know more.

Etot = Ep + Ek

We can show the variation of the energy graphically:

Now we will look at the energy with displacement:

If you are not sure about this, the key points to remember are:

In the exam, you will probably be asked simply to mark on a diagram where the maximum potential and kinetic energy occur.