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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

Radhika Batra , 11 Years ago
Grade 11
anser 3 Answers
Shane Macguire

Last Activity: 11 Years ago

mukunda

Last Activity: 11 Years ago

hii   ans is D(0.1256).     bye.         

Kushagra Madhukar

Last Activity: 4 Years 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

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