Question

For a particle performing linear S.H.M. of amplitude 'r', the potential energy is '\(λ\)' times its total energy. The displacement of particle is

• A. \(rλ\)
• B. \(\frac{r}{λ}\)
• C. \(r\sqrt{λ}\)
• D. \(\frac{r}{\sqrt{λ}}\)

Hint:

A slow clock has too large a time period, so the pendulum length must be reduced.

Answer 1

The Correct Option is C

Step 1: Relation:
$T\propto\sqrt l$.

Step 2: Apply:
To bring the period down to its correct value, shorten the pendulum (C).

Final Answer:
Reduce the length. \[ \boxed{\text{(C)}} \]