JEE MAIN - Physics (2007 - No. 24)
The displacement of an object attached to a spring and executing simple harmonic motion is given by $$x = 2 \times {10^{ - 2}}$$ $$cos$$ $$\pi t$$ metre. The time at which the maximum speed first occurs is
$$0.25$$ $$s$$
$$0.5$$ $$s$$
$$0.75$$ $$s$$
$$0.125$$ $$s$$
Explanation
Here, $$x = 2 \times {10^{ - 2}}\cos \,\pi \,t$$
$$\therefore$$ $$v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t$$
For the first time, the speed to be maximum,
$$\sin \pi t = 1$$ or, $$\sin \pi t = \sin {\pi \over 2}$$
$$ \Rightarrow \pi t = {\pi \over 2}\,\,\,$$ or, $$\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$$
$$\therefore$$ $$v = {{dx} \over {dt}} = 2 \times {10^{ - 2}}\,\pi \sin \pi t$$
For the first time, the speed to be maximum,
$$\sin \pi t = 1$$ or, $$\sin \pi t = \sin {\pi \over 2}$$
$$ \Rightarrow \pi t = {\pi \over 2}\,\,\,$$ or, $$\,\,\,\,t = {1 \over 2} = 0.5\,\sec .$$
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