A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of $$4$$ cm. The time period of the motion is $$1$$ s. At $$t=0$$, the cork is at its lowest position of oscillation, the position and velocity of the cork at t $$=$$ $$10.5$$ s, would be

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Infinity Learn · Accepted Answer

As the range of motion is $$4$$ cm, the amplitude of motion is $$+2$$ cm. $$10.5$$ s is equal to $$10.5$$ time period of simple harmonic motion; so we have to find the height and position of cork after $$one-half$$ of time period as $$T/2$$ $$=$$ $$0.5$$ s. As at t $$=$$ $$10$$ s, the particle is at its lowest position, after half a time period the cork would be at its maximum height and velocity of cork at extreme position is zero.

A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of 4 cm. The time period of the motion is 1 s. At t=0, the cork is at its lowest position of oscillation, the position and velocity of the cork at t = 10.5 s, would be

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