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# A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3.92 × 10–3 m. What must be the least period of these oscillations, so that the object is not detached from the platform?    (A)    0.1356 sec    (B)    0.1456 sec    (C)    0.1556 sec    (D)    0.1256 sec

Shane Macguire
30 Points
7 years ago

mukunda
33 Points
7 years ago

hii   ans is D(0.1256).     bye.

askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem below.

We know, max acceleration during SHM = amax = w2A = (2π/T)2 x A
Hence,T = 2π√(A/amax)
Now, amax = g = 9.81
Hence, T = √(3.92 x 10-3/9.81)
Hence, T = 0.12553 s
Hence correct answer is (d) 0.1256 s

Hope it helps.
Thanks and regards,
Kushagra