Question

A particle executing simple harmonic motion with amplitude A has the same potential and kinetic energies at the displacement

• A. \(2\sqrt{A}\)
• B. \(\frac{A}{2}\)
• C. \(\frac{A}{\sqrt{2}}\)
• D. \(A\sqrt{2}\)

Answer 1

The Correct Option is C

To solve this problem, we need to understand the conditions under which the potential energy (PE) and kinetic energy (KE) of a particle in simple harmonic motion (SHM) are equal. Simple Harmonic Motion is described by the equation:

• \(x(t) = A \cos(\omega t + \phi)\), where \(A\) is the amplitude. 

In SHM, the total mechanical energy \(E\) is constant and given by:

• \(E = \text{KE} + \text{PE}\).

The potential energy in SHM is given by:

• \(\text{PE} = \frac{1}{2} k x^2\), where \(k\) is the spring constant, and \(x\) is the displacement from the mean position.

The kinetic energy is given by:

• \(\text{KE} = \frac{1}{2} m \omega^2 (A^2 - x^2)\), where \(m\) is the mass of the particle, and \(\omega\) is the angular frequency.

At the displacement where potential and kinetic energy are equal, we have:

• \(\text{PE} = \text{KE}\)
• This implies: \(\frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)\)

Since \(\omega^2 = \frac{k}{m}\), substitute this and simplify:

• \(k x^2 = k (A^2 - x^2)\)
• This simplifies to: \(2 x^2 = A^2\)
• Therefore, \(x^2 = \frac{A^2}{2}\)
• Thus, \(x = \frac{A}{\sqrt{2}}\)

Thus, the displacement at which potential and kinetic energies are equal is \(\frac{A}{\sqrt{2}}\).

Therefore, the correct option is \(\frac{A}{\sqrt{2}}\).