Question

The amplitude of a particle executing SHM is $3 \,cm$ The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is:_______ $cm$

Answer 1

Correct Answer: 2

To solve this problem, we need to determine the displacement \( x \) at which the kinetic energy (KE) of the particle is 25% more than its potential energy (PE) in simple harmonic motion (SHM).

In SHM, the expressions for kinetic and potential energy are:
1. \( KE = \frac{1}{2}m\omega^2(A^2 - x^2) \)
2. \( PE = \frac{1}{2}m\omega^2x^2 \)
where \( A \) is the amplitude, \( x \) is the displacement, \( \omega \) is the angular frequency, and \( m \) is the mass of the particle.

We need to find \( x \) such that:
\( KE = 1.25 \, PE \)
Substitute the expressions:
\( \frac{1}{2}m\omega^2(A^2 - x^2) = 1.25 \times \frac{1}{2}m\omega^2x^2 \)
Cancel \(\frac{1}{2}m\omega^2\) from both sides:
\( A^2 - x^2 = 1.25x^2 \)
Rearrange and combine like terms:
\( A^2 = 2.25x^2 \)
Solve for \( x^2 \):
\( x^2 = \frac{A^2}{2.25} \)
Given \( A = 3 \, \text{cm} \), substitute to find \( x \):
\( x^2 = \frac{3^2}{2.25} = \frac{9}{2.25} = 4 \)
Take the square root to find \( x \):
\( x = \sqrt{4} = 2 \, \text{cm} \)
Verify the range: The calculated displacement \( 2 \, \text{cm} \) falls within the specified range (2, 2).
Thus, the displacement at which the kinetic energy is 25% more than the potential energy is \( \mathbf{2 \, cm} \).