A particle executes S.H.M. of amplitude A. If $$I1$$, and $$T2$$, are the times taken by the particle to traverse from $$0$$ to $$A/2$$ and from A I $$2$$ to A respectively, $$then(T1$$ $$/$$ $$T2)$$ will be equal to

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A particle executes S.H.M. of amplitude A. If $$I1$$, and $$T2$$, are the times taken by the particle to traverse from $$0$$ to $$A/2$$  and from A I $$2$$ to A respectively, $$then(T1$$ $$/$$ $$T2)$$ will be equal to

Infinity Learn · Accepted Answer

For S.H.M.$$x=Asin$$⁡ω$$t=Asin$$⁡(2$$$$π$$$$t/T) he time taken by the particle for travelling from x $$=$$ $$0$$ to x x $$=Al2isTt$$.Hence $$A2=Asin$$⁡$$2$$$$π$$$$T1T$$ or $$sin$$⁡$$2$$$$π$$$$T1T=12$$ or $$2$$$$π$$$$T1T=$$π$$2$$  or  $$T1=T12$$ Similarly, time taken bythe particle to traverse from x $$=0$$ to x $$=$$,$$4$$ i.e., $$one-fourth$$ cycle is $$T4=T1+T2$$ $$T2=T4$$−$$T1=T4$$−$$T12=T6$$  From eqs. (1) and (2), we get  $$T1T2=(T/12)T/6=12$$

A particle executes S.H.M. of amplitude A. If I₁, and T₂, are the times taken by the particle to traverse from 0 to A/2 and from A I 2 to A respectively, then(T₁ / T₂) will be equal to

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A particle executes S.H.M. of amplitude A. If I1, and T2, are the times taken by the particle to traverse from 0 to A/2 and from A I 2 to A respectively, then(T1 / T2) will be equal to

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A particle executes S.H.M. of amplitude A. If I₁, and T₂, are the times taken by the particle to traverse from 0 to A/2 and from A I 2 to A respectively, then(T₁ / T₂) will be equal to