Question

A simple pendulum consisting of a point mass 'm' tied to a massless string of length 'l' executes small oscillations of frequency 'f' and amplitude A=lx. What is the average (over a complete time period T of the pendulum) tension of the string?​

Answer 1

Answer:

T=mg because string is massless

Answer 2

Given that,

Mass = m

Length = l

Amplitude = a

The string has maximum tension when the bob is at mean position of oscillation.

Maximum amplitude of oscillation is a.

According to figure,

$BP=a$

$OP=\sqrt{l^2-a^2}$

$AP=OA-OP$

$AP=l-\sqrt{l^2-a^2}$

The kinetic energy at position A and the potential energy at point B are conserved.

We need to calculate the average tension of the string

Using conservation of energy

$\dfrac{1}{2}mv^2=mg(l-\sqrt{l^2-a^2})$

$v^2=2g(l-\sqrt{l^2-a^2})$

Using equation of circular motion at A

$\dfrac{2mg(l-\sqrt{l^2-a^2})}{l}=T-mg$

$T+mg+2mag(1-\sqrt{1-\dfrac{a^2}{l^2}})$

Using approximation

$\sqrt{1-x^2}=1-\dfrac{x^2}{2}$ for x<<1

$T=mg+mg(\dfrac{a}{l})^2$

$T=mg(1+\dfrac{a^2}{l^2})$

Hence, The average tension of the string is $mg(1+\dfrac{a^2}{l^2})$