Question

A simple pendulum with length $L$ and mass of the bob is vibrating with an amplitude $a$. Then the maximum tension in the string is:

• A. $mg$
• B. $mg\left[{1+\left(\frac{a}{L}\right)}^2\right]$
• C. $mg{[1+\frac{a}{2L}]}^2$
• D. $mg{[1+\left(\frac{a}{L}\right)]}^2$

Answer 1

The correct option is B $mg\left[{1+\left(\frac{a}{L}\right)}^2\right]$

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

Maximum amplitude of oscillation is $a$ i.e. $BP=a$

From geometry, $OP=\sqrt{L^2-a^2}$

Also, $AP=OA-OP=L-\sqrt{L^2-a^2}$

The whole kinetic energy of bob at position A is converted into its potential energy at position B.

$\therefore$ $\frac{1}{2}m{v}^2=mg(L-\sqrt{L^2-a^2})$ $⟹v^2=2g(L-\sqrt{L^2-a^2})$

Using circular motion equation at A :

$\frac{2mg(L-\sqrt{L^2-a^2})}{L}=T-mg$

$\therefore$ $T=mg+2mag(1-\sqrt{1-\frac{a^2}{L^2}})$

Using approximation $\sqrt{1-x^2}=1-\frac{x^2}{2}$ for $x<<1$

We get $T=mg+2mag(1-[1-\frac{a^2}{2L^2}])=mg+mg(\frac{a}{L}{)}^2$

$⟹$ $T=mg[1+\frac{a^2}{L^2}]$