Question

A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E . At one instant its kinetic energy is 3E/4 . Its displacement at that instant is

A. a/√2
B. a/2
C. a√3/2
D. a/√3

Answer 1

Correct option: A. a/√2. The displacement of the particle when its kinetic energy is 3E/4 in simple harmonic motion is a/2, calculated using the conservation of energy and the relationship between kinetic and potential energies in SHM.

Step-by-step explanation:

In simple harmonic motion (SHM), the total energy E of the system is conserved and shared between kinetic energy and potential energy.

At the mean position, all energy is kinetic, and at maximum displacement (amplitude a), all energy is potential.

The relationship between kinetic energy (KE) and potential energy (PE) at any point in the oscillation is given by E = KE + PE.

When the particle has a kinetic energy of 3E/4, the remaining 1E/4 must be potential energy, since total energy E is conserved.

The potential energy in SHM is proportional to the square of the displacement from the mean position and can be represented as PE = (kx^2)/2, with k being the spring constant and x the displacement.

Given that PE = 1E/4 and the total energy E = kA^2/2, we can set up the equation (kx^2)/2 = (kA^2/8).

Solving for x, we find that x^2 = A^2/4, therefore x = A/2.

Thus, the displacement of the particle when its kinetic energy is 3E/4 is a/2.