Question

A particle is executing simple harmonic motion with amplitude \( A \). The ratio of the kinetic energies of the particle when it is at displacements of \( \frac{A}{4} \) and \( \frac{A}{2} \) from the mean position is:

• A. 4 : 1
• B. 2 : 1
• C. 5 : 4
• D. 9 : 16

Hint:

For SHM, the kinetic energy depends on the displacement from the equilibrium position, and you can calculate it using the formula \(KE = \frac{1}{2} m \omega^2 (A^2 - x^2)\).

Answer 1

The Correct Option is A

In simple harmonic motion, the kinetic energy at a displacement \(x\) is given by: \[ KE = \frac{1}{2} m \omega^2 \left(A^2 - x^2\right), \] where \(A\) is the amplitude, \(x\) is the displacement, and \(\omega\) is the angular frequency. For the displacement \(x = \frac{A}{4}\) and \(x = \frac{A}{2}\), the kinetic energies are: \[ KE_1 = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{4}\right)^2\right), \] \[ KE_2 = \frac{1}{2} m \omega^2 \left(A^2 - \left(\frac{A}{2}\right)^2\right). \] The ratio of the kinetic energies is: \[ \frac{KE_1}{KE_2} = \frac{A^2 - \left(\frac{A}{4}\right)^2}{A^2 - \left(\frac{A}{2}\right)^2} = \frac{4}{1}. \]