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: Understanding the Concept:
Time period of a simple pendulum \(T = 2\pi\sqrt{\frac lg}\). It depends on length and g, not on mass or (for small swings) amplitude.

Step 2: Interpret:
A slow clock takes longer for each oscillation, so its period is too large. We need to reduce the period.

Step 3: Decide:
Reduce the length, since \(T\propto\sqrt l\). Changing mass does nothing, and reducing amplitude has almost no effect for small swings. Increasing length makes the clock even slower.

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