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. L/2
• B. 13L/16
• C. 15L/16
• D. 3L/4

Answer 1

The correct option is C

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)

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

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

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