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: Set up the problem.
A body oscillates with period $T_1$ under restoring force $F_1$, and with period $T_2$ under force $F_2$. Both forces are then applied together in the same direction, and we want the resulting period $T$.
Step 2: Link force to force constant.
Each linear restoring force has the form $F = -kx$. The period is $T = 2\pi\sqrt{\dfrac{m}{k}}$, so $k = \dfrac{4\pi^2 m}{T^2}$.
Step 3: Add the forces.
Acting together, $F_{net} = -k_1 x - k_2 x = -(k_1 + k_2)x$, so the effective constant is $k_{eq} = k_1 + k_2$.
Step 4: Write each constant via its period.
$k_1 = \dfrac{4\pi^2 m}{T_1^2}$, $k_2 = \dfrac{4\pi^2 m}{T_2^2}$, $k_{eq} = \dfrac{4\pi^2 m}{T^2}$.
Step 5: Substitute into $k_{eq} = k_1 + k_2$.
Cancelling the common $4\pi^2 m$, $\dfrac{1}{T^2} = \dfrac{1}{T_1^2} + \dfrac{1}{T_2^2} = \dfrac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.
Step 6: Invert and take the root.
$T^2 = \dfrac{T_1^2 T_2^2}{T_1^2 + T_2^2}$, so $T = \dfrac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$, option (4).
\[ \boxed{T = \dfrac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}} \]