FORCES ACTING ON THE BOB 


1. Weight of the bob (W) acting vertically downward. 2. Tension in the string (T) acting along the string. 

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

a. Wcosq along the string. b. Wsinq perpendicular to string. 

Since there is no motion along the string, therefore, the component Wcosq must balance tension (T) i.e. Wcosq = T 

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

According to Newton’s second law of motion Wsinq will be equal to m x a 

i.e. Wsinq = m a 

Since Wsinq is towards the mean position, therefore, it must have a negative sign. i.e. m a = – Wsinq 
But W = mg 
m a = – mgsinq 
a = – gsinq 
In our assumption q is very small because displacement is small, in this condition we can take sinq = q 
Hence a = – gq ———– (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. 