First slide

Simple hormonic motion

Question

In a S.H.M. when the displacement is one half the amplitude, what fraction of the total energy is kinetic ?

Easy

A

zero
B

1 /4
C

1 /2
D

3 /4

Solution

$In case of S.H.M. K . E . = \frac{1}{2} {mω}^{2} (a^{2} - x^{2}) ∴ K.E. = \frac{1}{2} {mω}^{2} [a^{2} - (a / 2)^{2}], (when x = a / 2) = \frac{3}{4} [\frac{1}{2} {mω}^{2} a^{2}] total energy = \frac{1}{2} {mω}^{2} a^{2} raction of total energy = \frac{K.E.}{Total energy} = \frac{\frac{3}{4} [\frac{1}{2} {mω}^{2} a^{2}]}{\frac{1}{2} {mωa}^{2}} = \frac{3}{4}$

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