Question

Maximum tension in a simple pendulum oscillating with amplitude $a$ and length $L$ is

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

Answer 1

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

$\text{Given:-}$ Lenght = L

Mass of the bob = M

Amplitude = a

$\text{To find:-}$ Maximum tension in the string

$\text{Solution:-}$

The tension in the string would be maximum when bob will be passing mean position.

$T_{max}=mg+\frac{m{v}^2}{L}$

At mean position

$v=a\omega$ and

$\omega =\sqrt{\frac{g}{L}}$

$\Rightarrow v=a\sqrt{\frac{g}{L}}$

$T_{max}=mg+\frac{m}{L}\left(\frac{a^{2}g}{L}\right)$

$T_{max}=mg[1+\frac{a^2}{L^2}]$

$\text{Hence the correct optiom is B}$