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Questions  

A body executing SHM has a maximum acceleration equal to 48 m/sec2 and maximum velocity equal to 12 m/sec. The amplitude of SHM is

a
3 m
b
332 m
c
10249 m
d
649 m

detailed solution

Correct option is A

given maximum acceleration amax=48m/s2Maximum velocity  Vmax=12m/samplitude A=? here angular velocity =ωamaxVmax=ω2AωA=4812=4=> ω=4 Vmax=ωAsubstitute maximum velocity, and angular velocity12=4(A)A=3m.

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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 3.92×103m . What must be the least period of these oscillations, so that the object is not detached from the platform 


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