A body executes SHM under the action of force ' F₁ ' with time period ' T₁ '. If the force is changed to ' F₂ ', it executes SHM with period ' T₂ '. If both the forces ' F₁ ' and ' F₂ ' act simultaneously in the same direction on the body, its time period is (a) (√(T₁²-T₂²))/(T₁T₂) (b) (T₁T₂)/(√(T₁²-T₂²)) (c) (√(T₁²+T₂²))/(T₁T₂) (d) (T₁T₂)/(√(T₁²+T₂²))

Answer: (d). Forces add: k = k₁ + k₂, and k ∝ 1/T². 1/T² = (1)/(T₁²) + (1)/(T₂²) T = (T₁T₂)/(√(T₁²+T₂²)), so the answer is (d).

A body executes SHM under the action of force ' F₁ ' with time period… | MHT CET 2021 (23 Sep, Shift 2)

Q71MHT CET 2021 · 23 SEP · SHIFT 2Medium

Spring Systems

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

Show answer and solution

Answer: (d)

Forces add: $k = k_1 + k_2$ , and $k \propto \frac{1}{T^2}$ .
$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$
$T = \frac{T_1T_2}{\sqrt{T_1^2+T_2^2}}$ , so the answer is (d).

Report a mistake

Spring Systems question from MHT CET 2021 (23 Sep, Shift 2)

More questions on Spring Systems