Physics - SHM Question with Solution | TestHub

The maximum tension in the string of a simple pendulum occurs at the mean position (lowest point of the oscillation).

The equation for tension at the mean position is given by:

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

For a simple pendulum executing simple harmonic motion, the maximum velocity is $v_{max}=\omega A$, where $\omega$ is the angular frequency and $A$ is the amplitude.

The angular frequency of a simple pendulum is $\omega =\sqrt{\frac{g}{L}}$.

Therefore, the square of the maximum velocity is:

$v_{max}^2={\omega }^2{A}^2=\frac{g}{L}{A}^2$

Substituting $v_{max}^2$ into the tension equation:

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

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

$T_{max}=mg\left[1+{\left(\frac{A}{L}\right)}^2\right]$

Answer: $mg\left[1+{\left(\frac{A}{L}\right)}^2\right]$