Question

- A particle is undergoing SHM, when its kinetic
energy is equal to potential energy, the velocity of
particle is (w - angular frequency, A - amplitude
and potential energy at mean position is zero)
​

Answer 1

1/2mv^2 = 1/2kx^2

v =√A^2 - x^2

$\frac{1}{2} m( {w}^{2} ( {a}^{2} - {x}^{2} )) = \frac{1}{2} k {x}^{2}$

solve for x

Answer 2

Answer:

$V=\dfrac{\omega A}{\sqrt 2}}$

Explanation:

 We know that

Velocity of particle in SHM given as

$V=\omega \sqrt{A^2-x^2}$

Kinetic energy of particle given as

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}m\times \omega^2 (A^2-x^2)$

Potential energy given as

$PE=\dfrac{1}{2}kx^2$

Given that

KE= PE

$\dfrac{1}{2}m\times \omega^2 (A^2-x^2)=\dfrac{1}{2}kx^2$

$m\times \omega^2 (A^2-x^2)=kx^2$

We know that

ω² m = k

So

$(A^2-x^2)=x^2$

$x=\dfrac{A}{\sqrt 2}$

Now

$V=\omega \sqrt{A^2-x^2}$

$V=\omega \sqrt{A^2-\dfrac{A^2}{2}}$

$V=\omega \sqrt{\dfrac{A^2}{2}}$

$V=\dfrac{\omega A}{\sqrt 2}}$

This is the velocity of particle when  when its kinetic  energy is equal to potential energy.