Question

A particle is undergoing SHM, when its kinetic
energy is equal to potential energy, the velocity of
particle is (0 - angular frequency, A - amplitude
and potential energy at mean position is zero)​

Answer 1

The velocity of particle is $\frac{\omega _{o}A}{\sqrt{2}}$.

Explanation:

1. Angular frequency = $\omega _{o}$ = $\sqrt{\frac{k}{m}}$   ...1)

2. Amplitude = A

3. during SHM, total energy is conserved.

  Means

  Kinetic energy + Potential Energy = Maximum Kinetic energy = Maximum Potential Energy   ...2)

 It is given that at certain instant

  Kinetic energy  = Potential Energy   ...3)

4. from equation 2) and equation 3)

   Kinetic energy + Kinetic energy = Maximum Potential Energy

   2×Kinetic energy = Maximum Potential Energy   ...4)

5. Where

    Kinetic energy = $\frac{1}{2}\times m\times V^{2}$    ...5)

    where V = velocity of particle at that instant

    Maximum Potential Energy=  $\frac{1}{2}\times k\times A^{2}$   ...6)

7. So equation 4) can be written as

   $2\times \frac{1}{2}\times m\times V^{2}=\frac{1}{2}\times k\times A^{2}$

  So

    $V =(\sqrt{\frac{k}{2m}})\times A$     ...7)

8.   From equation 1) and equation 7) can be written as

     $V =\frac{\omega _{o}A}{\sqrt{2}}$