Questions
A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4cm from the mean position. The magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time in seconds is
detailed solution
Correct option is A
Amplitude A=5cm given Velocity of particle=acceleration of particle at displacement y=4 ⇒ωA2−y2=ω2y, here w is angular velocity ⇒52−42=ω.4 34=ω=2πT ⇒time period is T=8π3sTalk to our academic expert!
Similar Questions
A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is . What must be the least period of these oscillations, so that the object is not detached from the platform
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