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PhysicsSHMEnergyHard2 minPYQ_2023

PhysicsHardsingle choice

A particle executes SHM of amplitude $A$ . The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

Options:

Answer:

Solution:

The potential energy is given by
$U = \frac{m ω^{2} x^{2}}{2}$
The kinetic energy is given by $K = \frac{m ω^{2} (A^{2} - x^{2})}{2}$ .
Using the condition that $K = U$ ,
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$\Rightarrow x = \frac{A}{\sqrt{2}}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: PYQ_2023

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