The extension of a spring is directly proportional to the force (Hooke’s Law). Consider a mass, m, put onto a spring of spring constant k so that so that it stretches by an extension l.
force on the spring =
and the stretching tension =
Ž mg = kl
Why does the spring stretch when you add a mass to it?
the spring is pulled down by a distance
below the rest position. Now the
stretching force become
This is also the tension in the spring acting upwards.
So the restoring force,
– mg. This is
is the weight, which always acts downwards.
we can write:
Fup = kl + kx – kl = kx
We can now apply Newton II to write:
-kx = ma
sign tells us that the force is upwards)
We know from SHM that
)2 x, so we can
So we can tidy this up to say that:
Since a = - (2pf )2 x, we can say that the condition for SHM is satisfied in this system, as long as Hooke’s Law is obeyed. We can now rearrange the equation above to write:
This then becomes:
we can now write down an expression to relate the period with the mass and the
spring constant :
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
which we can approximate to
This is a part of a required practical.
The relationship of the graph suggests that the line should cut through the origin. However we may 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.
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 9.8 m s-2?
need to work out the spring constant using Hooke’s Law
9.8 m s-2 = 4.9 N m-1
we can use
T = 2pÖ(m/k)
to work out the period:
4.9 N m-1
A common bear trap is to forget to take the square root.
A spring has a spring constant of 80 N m-1. 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. We need the angle to be small, so that we can say that the arc OA is (nearly) the same length as the chord OA.
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:
The m terms cancel out. Therefore:
The relationship a = -(2pf )2s arises because this is a simple harmonic oscillator.
In this case the s terms cancel out to give:
This rearranges to:
This then becomes:
We know that period is given by:
So we turn the equation upside down and write the
formula linking period,
T, with length,
l, and gravity constant,
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. Note that we have not used the angular velocity term w. It is easier to use 2pf.
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-1, compared with 9.8 N kg-1 on earth.
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. The graph is like this:
We can get a relatively accurate determination of this if we:
Count at least 100 swings
Use a swing angle of less than 10o.
Measure l to the centre of the bob.
Count the oscillations as the bob passes the centre position.
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 = 5.0 cm.
We need to know w.
w = 2p/T = 2p rad ø 2.0 s = p rad s-1
Now we can work out the velocity:
v2 = p2(5.02 – 02) = p2 ´ 25 cm = 246.74 cm2 s-2
Ž v = 15.7 cm s-1
We can use a = -w2s. Acceleration is at a maximum when s = A = 5.0 cm
Ž a = -(p rad s-1)2 ´ 5.0 cm = - 49.3 cm s-2. The negative sign tells us that the direction of the acceleration is to the rest position.
What is the time period for a pendulum of length 4.6 m. Take g = 9.8 m s-2
The amplitude of the swing of the pendulum in question 5 is 0.50 m. What is the maximum acceleration? What is the maximum velocity?
Energy in Simple Harmonic
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:
Potential energy is highest when the oscillator is at the maximum amplitude;
Kinetic energy is highest when the oscillator passes the rest position.
In the exam, you will probably be asked simply to mark on a diagram where the maximum potential and kinetic energy occur.