Questions
The velocity of particle undergoing SHM at the mean position is 4m/s. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude
detailed solution
Correct option is A
maximum velocity of simple harmonic oscillator is Vmax=w×A where w=angular velocity, A=amplitude, given, wA=4m/s we know that, Velocity of oscillator is V=wA2-x2 here x= displacement of the oscillator ⇒V=wA2-A2/4 ⇒V=w3A2/4 ⇒V=w3A2, substitute wA=4 v=23m/sTalk 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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