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Question 13

A particle is executing simple harmonic motion with a time period $$T$$. At time $$t = 0$$, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like:

For a particle in SHM,

$$x=A\sinωt$$

$$v=\frac{dx}{dt}=Aω\cosωt$$

$$KE=\frac{1}{2}mv^2=\frac{1}{2}mA^2ω^2\cos^2ωt=KE_{\max}\cos^2ωt$$

KE is zero when $$\cos^2ωt=0$$, i.e. $$ωt=\frac{\pi}{2},\ \frac{3\pi}{2},...$$ or $$t=\frac{T}{4},\ \frac{3T}{4},...$$

At t=0, $$KE=KE_{\max}$$

KE is always positive

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