Question

A body is executing SHM. When its displacements from the mean position are 4 cm and 5 cm, it has velocity 10 cm/s and 8 cm/s respectively. Its periodic time $ t $ is

• A. \( \frac{2\pi}{3} \) sec
• B. \( 2\pi \) sec
• C. \( \frac{3\pi}{2} \) sec
• D. \( \pi \) sec

Hint:

To calculate the periodic time in SHM, use the relationship between velocity, displacement, and angular frequency. Solve for angular frequency and then find the time period.

Answer 1

The Correct Option is D

In Simple Harmonic Motion (SHM), the velocity \( v \) is related to the displacement \( x \) and the amplitude \( A \) by the equation: \[ v = \omega \sqrt{A^2 - x^2} \] Where: - \( \omega \) is the angular frequency, - \( x \) is the displacement at any point in time, - \( A \) is the amplitude of the motion. From the given data: - When \( x = 4 \, \text{cm} \), \( v = 10 \, \text{cm/s} \), - When \( x = 5 \, \text{cm} \), \( v = 8 \, \text{cm/s} \). We can use these two pieces of information to form two equations and solve for \( \omega \) and \( A \). For \( x = 4 \, \text{cm} \) and \( v = 10 \, \text{cm/s} \): \[ 10 = \omega \sqrt{A^2 - 16} \] For \( x = 5 \, \text{cm} \) and \( v = 8 \, \text{cm/s} \): \[ 8 = \omega \sqrt{A^2 - 25} \] Now, solving these equations: From the first equation: \[ \omega = \frac{10}{\sqrt{A^2 - 16}} \] Substitute into the second equation: \[ 8 = \frac{10}{\sqrt{A^2 - 16}} \cdot \sqrt{A^2 - 25} \] Simplifying, we find \( A \) and \( \omega \), which allows us to calculate the periodic time \( T = \frac{2\pi}{\omega} \). After solving, we find that the periodic time \( T = \pi \) sec.