Question

A body performs S.H.M. under the action of force '$F_1$' with period '$T_1$' second. If the force is changed to '$F_2$' it performs S.H.M. with period '$T_2$' second. If both forces '$F_1$' and '$F_2$' act simultaneously in the same direction on the body, the period in second will be

• A. $\frac{T_1 + T_2}{T_1 T_2}$
• B. $\frac{T_1^2 + T_2^2}{T_1 T_2}$
• C. $\frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$
• D. $\frac{T_1 T_2}{T_1 + T_2}$

Hint:

This scenario is mathematically identical to connecting two springs in parallel. For parallel springs, $K_{\text{eq}} = K_1 + K_2$, which always leads to the time period relationship $1/T^2 = 1/T_1^2 + 1/T_2^2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A body undergoes SHM under two different restoring forces individually, yielding two different time periods. We need to find the new time period when both forces act on it simultaneously in the same direction.

Step 2: Key Formula or Approach:
The restoring force in SHM is $F = -Kx$, where $K$ is the force constant.
The time period is given by $T = 2\pi\sqrt{\frac{m}{K}}$, which can be rearranged to $K = \frac{4\pi^2 m}{T^2}$.
When both forces act together in the same direction, the effective force constant is $K_{\text{eff}} = K_1 + K_2$.

Step 3: Detailed Explanation:
Let the mass of the body be $m$.
For force $F_1 = -K_1 x$:
$$T_1 = 2\pi\sqrt{\frac{m}{K_1}} \implies K_1 = \frac{4\pi^2 m}{T_1^2}$$
For force $F_2 = -K_2 x$:
$$T_2 = 2\pi\sqrt{\frac{m}{K_2}} \implies K_2 = \frac{4\pi^2 m}{T_2^2}$$
When both forces act simultaneously, the net force is $F_{\text{net}} = F_1 + F_2 = -(K_1 + K_2)x$.
The effective force constant is $K = K_1 + K_2$.
The new time period $T$ is given by:
$$T = 2\pi\sqrt{\frac{m}{K_{\text{eff}}}}$$
Squaring both sides and reciprocating:
$$\frac{1}{T^2} = \frac{K_{\text{eff}}}{4\pi^2 m} = \frac{K_1 + K_2}{4\pi^2 m}$$
Substitute $K_1$ and $K_2$:
$$\frac{1}{T^2} = \frac{\frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}}{4\pi^2 m}$$
Cancel out $4\pi^2 m$:
$$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$$
Find a common denominator:
$$\frac{1}{T^2} = \frac{T_2^2 + T_1^2}{T_1^2 T_2^2}$$
Invert both sides to solve for $T^2$, and then take the square root:
$$T^2 = \frac{T_1^2 T_2^2}{T_1^2 + T_2^2} \implies T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$$

Step 4: Final Answer:
The combined time period is $\frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$, matching option (C).