SIMPLE PENDULUM

Simple pendulum consists of a heavy mass particle suspended by a light, flexible and in-extensible string.

MOTION OF THE BOB OF SIMPLE PENDULUM: -

The motion of the bob of simple pendulum, simple harmonic motion if it is given small displacement. In order to prove this fact considers a simple pendulum having a bob of mass 'm' and the length of pendulum is 'l'. Assuming that the mass of the string as pendulum is negligible. When the pendulum is at rest at position 'A', the only force acting is its weight and tension in the string. When it is displaced from its mean position to another new position say 'B' and released, it vibrates to and fro around its mean position.

   Suppose that at this instant the bob is at point 'B' as shown below:

FORCES ACTING ON THE BOB   : -

·         Weight of the bob (W) acting vertically downward.

·         Tension in the string (T) acting along the string.

The weight of the bob can be resolved into two rectangular components:

1.      W cos θ along the string.

2.      W sin θ perpendicular to string.

  Since there is no motion along the string, therefore, the component W cos θ must balance tension (T)

i.e.                                                                      W cos θ = T

  This shows that only W sin θ is the net force which is responsible for the acceleration in the bob of pendulum.

  According to Newton's second law of motion W sin θ will be equal to m × a

  i.e.                                                                      W sin θ = m a

  Since W sin θ is towards the mean position, therefore, it must have a negative sign.

i.e.                                                                     m a = -  W sin θ 

But W = mg

                                                                         m a = -  m g sin θ 

                                                                               a = - g sin θ  

In our assumption q is very small because displacement is small, in this condition we can take sinθ = θ

Hence                                                                     a = - g θ ----------- (1) 

If x be the linear displacement of the bob from its mean position, then from figure, the length of arc AB is nearly equal to x 

From elementary geometry we know that:

Where s = x, r = l

Putting the value of q in equation (1)

 As the acceleration of the bob of simple pendulum is directly proportional to displacement and is directed towards the mean position, therefore the motion of the bob is simple harmonic when it is given a small displacement.

Show that the motion of a mass attached to the end of a spring is SHM: -  

 

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed at the firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface. If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

 According to "Hook's Law 

F = - K x ---- (1)

   Negative sign indicates that the elastic restoring force is opposite to the displacement.

   Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves    forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given    by

F = m a ---- (2)

Comparing equation (1) & (2)

m a = - k x

Here k / m is constant term, therefore,

a = - (Constant) x

or

a ∞ -x

   This relation indicates that the acceleration of body attached to the end elastic spring is directly    proportional to its displacement. Therefore its motion is Simple Harmonic Motion.