Question

For small oscillations, a pendulum of length L is released from rest. Its length is L and time period is T. vertically below the point of suspension O, there is an obstacle P at a vertical distance x. the pendulum returns back to its original position after time $\frac{5T}{8}$. Find x.

• A. $\frac{L}{2}$
• B. $\frac{13L}{16}$
• C. $\frac{15L}{16}$
• D. $\frac{3L}{4}$

Answer 1

The correct option is C

$\frac{15L}{16}$

Time period $T=2\pi \sqrt{\frac{Length}{g}}\Rightarrow T\alpha (length{)}^{\frac{1}{2}}$.

After the pendulum hits the obstacle,the effective length becomes (L-x).Let the time period after

hitting the obstacle be T'.

$\therefore \frac{T^{\prime }}{T}=\sqrt{\frac{L-x}{L}}\Rightarrow T^{\prime }=T\sqrt{\frac{L-x}{L}}$ - - - - - - (1)

The total time taken to come back to initial position

$\frac{T}{2}+\frac{T^{\prime }}{2}=\frac{T}{2}+\frac{T}{2}\sqrt{\frac{L-x}{L}}=\frac{5T}{8}$ (as given)

$\Rightarrow \sqrt{\frac{L-x}{L}}=2(\frac{5}{8}-\frac{1}{2})=\frac{1}{4}$

$\Rightarrow \frac{L-x}{L}=\frac{1}{16}$ $\Rightarrow 15L=16x$

$\Rightarrow x=\left(\frac{15L}{16}\right)$