QJEEM23Ph3293: JEE Main 2023 Physics - Class 11 JEEM23Ph3293 Solution: A particle is executing simple harmonic motion (SHM). The ratio …

Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x. The given condition is that x = A/2.

For a particle in SHM, the potential energy (PE) is given by:

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

And the total mechanical energy (E) of the particle remains constant and is given by:

$E = \frac{1}{2} kA^2$

Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:

$E = PE + KE$

Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.

Calculate the potential energy at x = A/2:

$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$

Substitute the expression for total mechanical energy:

$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$

Now, find the ratio of potential energy to kinetic energy:

$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$

Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.