Question

A simple pendulum with lenght $l$ and bob of mass $m$ is executing S.H.M. of small amplitude $a$. The maximum tension in the string will be

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

Answer 1

The correct option is D $mg[1+(a/l{)}^2]$

Drawing the FBD of the pendulum at one extreme position,


Equation of motion along radial direction
$T-mgcos\theta =\frac{m{v}^2}{l}$
$T=\frac{m{v}^2}{l}+mgcos\theta$

$T$ will be maximum when $cos\theta =1$
$T_{max}=\frac{m{v}^2}{l}+mg$
$=\frac{m{a}^2{\omega }^2}{l}+mg$

Substituting ${\omega }^2=\frac{g}{l}$ (angular frequency for simple pendulum)

$T_{max}=\frac{m{a}^{2}g}{l^2}+mg=mg[1+{\left(\frac{a}{l}\right)}^2]$