First slide

Simple hormonic motion

Question

In simple harmonic motion how many times potential energy is equal to kienetic energy during one complete period?

Moderate

A

1
B

2
C

8
D

4

Solution

If frequency of oscillation is f, then frequency of potential or kinetic energy is 2f.
Now, $K = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$ and $U = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$
When $K = U, \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$
$\Rightarrow tanωt = \pm 1, \Rightarrow ωt = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$
So in a cycle $K = U, at t = \frac{π}{4 ω}, \frac{3 π}{4 ω}, \frac{5 π}{4 ω}, \frac{7 π}{4 ω}$ .

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