A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be

<p>Let&#39;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.</p> <p>For a particle in SHM, the potential energy (PE) is given by:</p> <p>$PE = \frac{1}{2} kx^2$</p> <p>And the total mechanical energy (E) of the particle remains constant and is given by:</p> <p>$E = \frac{1}{2} kA^2$</p> <p>Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:</p> <p>$E = PE + KE$</p> <p>Now, we need to find the ratio of potential energy to kinetic energy when x = A/2. </p> <p>Calculate the potential energy at x = A/2:</p> <p>$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$</p> <p>Substitute the expression for total mechanical energy:</p> <p>$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$</p> <p>Now, find the ratio of potential energy to kinetic energy:</p> <p>$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$</p> <p>Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.</p>