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

Infinity Learn · Accepted Answer

Time taken by particle to move from x $$=$$ $$0$$ (mean$$ $$position) to x $$=$$ $$4$$ (extreme$$ $$position) $$=$$ $$T4$$ $$=$$ $$1.24$$ $$=$$ $$0.3$$ s Let t be the time taken by the particle to move from x $$=$$ $$0$$ to x $$=$$ $$2$$ cmy $$=$$ a $$sin$$ ωt   ⇒ $$2$$ $$=$$ $$4$$ $$sin2$$$$π$$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 $$=$$ $$2$$×$$0.2$$ $$=$$ $$0.4$$ $$sec$$

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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