A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

Question

Infinity Learn · Accepted Answer

Let displacement equation of particle executing SHM is y $$=$$ a $$sin$$ ωtAs particle travels half of the amplitude from the equilibrium position, so y $$=$$ $$a2Therefore$$, $$a2$$ $$=$$ a $$sin$$ ωtor $$sin$$ ωt $$=$$ $$12=$$ $$sin$$ π$$6$$ or ωt $$=$$ π$$6or$$ t $$=$$ π$$6$$ω or t $$=$$ π$$6(2$$$$π$$$$T)(as$$ ω $$=$$ $$2$$$$π$$$$T)or$$ t $$=$$ $$T12Hence$$, the particle travels half of the amplitude from the equilibrium in $$T12sec$$

A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

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