First slide

Simple harmonic motion

Question

The total energy of a particle executing S.H.M. is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position

Moderate

A

60 J
B

10 J
C

40 J
D

45 J

Solution

$\frac{U}{E} = \frac{\frac{1}{2} {mω}^{2} y^{2}}{\frac{1}{2} {mω}^{2} a^{2}} = \frac{y^{2}}{a^{2}}$ $\Rightarrow \frac{U}{80} = \frac{{(\frac{3}{4} a)}^{2}}{a^{2}} = \frac{9}{16}$ $\Rightarrow U = 45 J$

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