Question

A particle starting from mean position oscillates simple harmonically with period 4 s. After what time will its kinetic energy be 75% of the total energy? (cos \( 30^\circ = \frac{\sqrt{3}}{2} \))

• A. \( \frac{1}{3} \, \text{s} \)
• B. \( \frac{1}{2} \, \text{s} \)
• C. \( \frac{1}{5} \, \text{s} \)
• D. \( \frac{1}{4} \, \text{s} \)

Hint:

In simple harmonic motion, the kinetic energy is proportional to the square of the displacement, and the time at which a certain kinetic energy is reached can be found by using the cosine function for displacement.

Answer 1

The Correct Option is A

Step 1: Kinetic Energy in Simple Harmonic Motion.
The total energy \( E \) of a particle in simple harmonic motion is given by: \[ E = \frac{1}{2} m \omega^2 A^2 \] where \( \omega \) is the angular frequency and \( A \) is the amplitude. The kinetic energy \( K \) is given by: \[ K = \frac{1}{2} m \omega^2 x^2 \] where \( x \) is the displacement from the mean position. The ratio of the kinetic energy to the total energy is: \[ \frac{K}{E} = \frac{x^2}{A^2} \] For \( K = 0.75 E \), we have: \[ \frac{x^2}{A^2} = 0.75 \quad \Rightarrow \quad \frac{x}{A} = \cos 30^\circ = \frac{\sqrt{3}}{2} \] Now, \( \cos \left( \omega t \right) = \frac{\sqrt{3}}{2} \), and solving for \( t \), we get: \[ t = \frac{1}{3} \, \text{s} \] Step 2: Final Answer.
Thus, the time after which the kinetic energy is 75% of the total energy is \( \frac{1}{3} \, \text{s} \).