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

The correct option is B

$Mg[1+(\frac{a}{L}{)}^2]$

Maximum tension in the string is at lowest position. Therefore, to find the velocity v at the lowest point of the path, we apply law of conservation of energy i.e.

$\frac{1}{2}M{v}^2=Mgh=MgL(1-cos\theta )$

$[\therefore h=L-Lcos\theta ]$

Or $v^2=2gL(1-cos\theta )$

Or $v=\sqrt{2gL(1-cos\theta )}$

$\therefore T=Mg+2Mg(1-cos\theta )$

$T=Mg[1+2\times 2si{n}^2(\frac{\theta }{2})]orT=Mg[1+4(\frac{\theta }{2}{)}^2]$

$[\because sin(\frac{\theta }{2})=\frac{\theta }{2}$ for small amplitudes $]$

$T=Mg[1+{\theta }^2]$

From figure $\theta =\frac{a}{L}$

$\therefore T=Mg[1+(\frac{a}{L}{)}^2]$