First slide

Simple hormonic motion

Question

A particle executing SHM with amplitude A has total energy E. At what distance from the mean position its kinetic energy will be $\frac{5 E}{9}$ ?

Easy

A

$\frac{A}{2}$
B

$\sqrt{\frac{2}{3}} A$
C

$\frac{3 A}{4}$
D

$\frac{2 A}{3}$

Solution

$E = \frac{1}{2} m ω^{2} A^{2} and K = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$
when $K = \frac{5 E}{9}, \frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{5}{9} \times \frac{1}{2} m ω^{2} A^{2}$
$\Rightarrow x^{2} = A^{2} - \frac{5}{9} A^{2} \Rightarrow x = \frac{2 A}{3}$

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