A particle executes linear simple harmonic motion with an amplitude of $$3$$ cm. When the particle is at $$2$$ cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Question

Infinity Learn · Accepted Answer

Given,  A $$=$$ $$3$$ cm, x $$=$$ $$2$$ cmThe velocity of a particle in simple harmonic motion is given $$asv=$$ω$$A2-x2$$ and magnitude of its acceleration is $$a=$$ω$$2x$$ Given |v|$$=$$|a|∴  ω$$A2-x2=$$ω$$2x$$ω$$x=A2-x2$$   or   ω$$2x2=A2-x2$$ω$$2=A2-x2x2=9-44=54$$ω$$=52$$ Time period, $$T=2$$πω$$=2$$π·$$25=4$$$$π$$$$5$$ s

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

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