Question

A body performs S.H.M. due to force \(F_1\) with time period \(0.8\,\text{s}\). If the force is changed to \(F_2\), it executes S.H.M. with time period \(0.6\,\text{s}\). Now both the forces act simultaneously in the same direction on the same body. The new periodic time is

• A. \(0.48\,\text{s}\)
• B. \(0.24\,\text{s}\)
• C. \(0.12\,\text{s}\)
• D. \(0.36\,\text{s}\)

Hint:

When multiple restoring forces act together, their force constants add up.

Answer 1

The Correct Option is A

Step 1: Relation between time period and force constant.
For SHM,
\[ T = 2\pi \sqrt{\frac{m}{k}} \Rightarrow k \propto \frac{1}{T^2} \]
Step 2: Determine individual force constants.
\[ k_1 \propto \frac{1}{(0.8)^2}, \quad k_2 \propto \frac{1}{(0.6)^2} \]
Step 3: Resultant force constant.
When both forces act together in the same direction,
\[ k = k_1 + k_2 \]
Step 4: Express new time period.
\[ \frac{1}{T^2} = \frac{1}{(0.8)^2} + \frac{1}{(0.6)^2} \] \[ \frac{1}{T^2} = \frac{1}{0.64} + \frac{1}{0.36} = 1.5625 + 2.7778 = 4.3403 \] \[ T = \frac{1}{\sqrt{4.3403}} \approx 0.48\,\text{s} \]
Step 5: Conclusion.
The new periodic time is \(0.48\,\text{s}\).