Question

A body executes SHM under the action of force '$F_1$' with time period '$T_1$'. If the force is changed to '$F_2$', it executes SHM with period '$T_2$'. If both the forces '$F_1$' and '$F_2$' act simultaneously in the same direction on the body, its time period is

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

Hint:

This formula matches the algebraic structure for combining parallel resistors or series capacitors: $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$. Whenever independent driving forces act together on a single mass, their rates add up, meaning the combined period is always smaller than either individual period.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A mass experiences an individual restoring force $F_1$ producing harmonic oscillations with period $T_1$, and a different restoring force $F_2$ yielding period $T_2$. We need to find the net period $T$ when both forces act together in the same direction on the identical mass.

Step 2: Key Formula or Approach:
For linear SHM, the restoring force is proportional to displacement: $F = -kx$, where $k$ is the force constant. The time period is:
$$T = 2\pi\sqrt{\frac{m}{k}} \implies T^2 = \frac{4\pi^2 m}{k} \implies k = \frac{4\pi^2 m}{T^2}$$ When both forces act together in the same direction:
$$F_{\text{net}} = F_1 + F_2 = -k_1 x - k_2 x = -(k_1 + k_2)x$$ This shows that the equivalent force constant is simply $k_{\text{eq}} = k_1 + k_2$.

Step 3: Detailed Explanation:
Let us express the individual spring constants in terms of their respective time periods:
$$k_1 = \frac{4\pi^2 m}{T_1^2}, \quad k_2 = \frac{4\pi^2 m}{T_2^2}, \quad k_{\text{eq}} = \frac{4\pi^2 m}{T^2}$$ Substitute these expressions into the spring constant addition equation ($k_{\text{eq}} = k_1 + k_2$):
$$\frac{4\pi^2 m}{T^2} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}$$ Dividing both sides by the common scalar factor $4\pi^2 m$:
$$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$$ Find a common denominator for the right-hand side:
$$\frac{1}{T^2} = \frac{T_2^2 + T_1^2}{T_1^2 T_2^2}$$ Inverting both sides of the equation to isolate $T^2$:
$$T^2 = \frac{T_1^2 T_2^2}{T_1^2 + T_2^2}$$ Taking the square root of both sides gives the composite time period:
$$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$$

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