Question

A small bob attached to a light inextensible thread of length $l$ has a periodic time $T$ when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end $\text{O}$ on a vertical rigid wall of length $\frac{3l}{4}$. If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be $\frac{NT}{4}$ where $N$ is equal to

Answer 1

The time period of the pendulum,
$T=2\pi \sqrt{\frac{l}{g}}$
Now the wall is introduced in the pendulum which will make pendulum to rotate about one end of wall for half of the cycle with the length of $(l-\frac{3l}{4}=\frac{l}{4})$ ​.
And in the remaining half cycle pendulum will rotate about the complete $l$ length. So we can divide the new time period $T_{new}$ in $T_1$ and $T_2$.
​
So, the new periodic time, $T_{new}$ will be equal to sum of their half time periods.
$\Rightarrow T_{new}=\frac{T_1}{2}+\frac{T_2}{2}$

$\Rightarrow T_{new}=\frac{1}{2}\times [2\pi \sqrt{\frac{l}{4g}}+2\pi \sqrt{\frac{l}{g}}]$

$\Rightarrow T_{new}=\frac{1}{2}\times \left[2\pi \sqrt{\frac{l}{g}}(1+\frac{1}{2})\right]$

$\Rightarrow T_{new}=\frac{1}{2}\times 2\pi \sqrt{\frac{l}{g}}\times \frac{3}{2}$

$\Rightarrow T_{new}=\frac{1}{2}\times T\times \frac{3}{2}$

$\therefore T_{new}=\frac{3T}{4}$

Accepted answer: 3, 3.0, 3.00