A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position

Question

A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

TestHub · Accepted Answer

The potential energy of a particle executing S.H.M. at a displacement $x$ is given by $U = \frac{1}{2} m ω^{2} x^{2}$ .

The maximum potential energy is $U_{max} = \frac{1}{2} m ω^{2} A^{2}$ .

Given that the potential energy is half of its maximum value:

$U = \frac{1}{2} U_{max}$

$\frac{1}{2} m ω^{2} x^{2} = \frac{1}{2} (\frac{1}{2} m ω^{2} A^{2})$

$x^{2} = \frac{A ^{2}}{2}$

$x = \pm \frac{A}{2}$

Answer: $\pm \frac{A}{2}$

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