If x,v and a denotes the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T. Then which of the following does not change with time?
A
a2T2+4π2v2
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B
aT/x
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C
aT+2πv
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D
aT/v
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Solution
The correct option is BaT/x We know that, a=−ω2x
(∵ particle executes SHM)
Muliplying on both sides with Tx we get : aTx=−ω2T (∵ω=2πT) ⇒aTx=−4π2T= constant
Since, the velocity of the particle is changing at every instant, therefore, every option other then option (b) changes with the time.