A particle is performing simple harmonic motion along $$x-axis$$ with amplitude $$4$$ cm and time period $$1.2$$ $$sec$$. The minimum time taken by the particle to move from x $$=2$$ cm to x $$=$$ $$+$$ $$4$$ cm and back again is given by

Question

A particle is performing simple harmonic motion along $$x-axis$$ with amplitude $$4$$ cm and time period $$1.2$$ $$sec$$. The minimum time taken by the particle to move from x $$=2$$ cm to x $$=$$ $$+$$ $$4$$ cm  and back again is given by

Infinity Learn · Accepted Answer

Time taken by particle to move from $$x=0$$ (mean$$ $$position) to x $$=$$ $$4$$ (extreme$$ $$position) $$=T4=1.24=0.3$$ sLet t be the time taken by the particle to move from $$x=0$$ to $$x=2$$ cm  $$y=asin$$ωt⇒$$2=4sin2$$$$π$$Tt⇒$$12=sin2$$$$π$$$$1.2t$$⇒π$$6=2$$$$π$$$$1.2t$$⇒$$t=0.1$$ s. Hence time to move from x $$=$$ $$2$$ to x $$=$$ $$4$$ will be equal to $$0.3$$ – $$0.1$$ $$=$$ $$0.2$$ sHence total time to move from x $$=$$ $$2$$ to x $$=$$ $$4$$ and back again $$=2$$×$$0.2=0.4sec$$

A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x =2 cm to x = + 4 cm and back again is given by

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