A body executes simple harmonic motion under the action of a force, F 1 with a time period 4 5 s . If the force is changed to F 2 , it executes SHM with time period 3 5 s . If both the forces F 1 and

Question

A body executes simple harmonic motion under the action of a force, F 1 with a time period 4 5 s . If the force is changed to F 2 , it executes SHM with time period 3 5 s . If both the forces F 1 and F 2 act simultaneously in the same direction on the body, its time period (in seconds) is

TestHub · Accepted Answer

F 1 = - k 1 x ; F 2 = - k 2 x a 1 = - k 1 m x ; a 2 = - k 2 m x Also, a 1 = - ω 1 2 x ; a 2 = - ω 2 2 x Now, resultant force F = F 1 + F 2 = - k 1 x - k 2 x - k x = - k 1 x - k 2 x k = k 1 + k 2 m ω 2 = m ω 1 2 + m ω 2 2 ω 2 = ω 1 2 + ω 2 2 2 π T 2 = 2 π T 1 2 + 2 π T 2 2 1 T 2 = 1 T 1 2 + 1 T 2 2 ∴ T = T 1 T 2 T 1 2 + T 2 2 ⇒ T = 4 5 × 3 5 4 5 2 + 3 5 2 = 12 25 s

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