Question

The frequency of a particle performing linear S.H.M. is \( \frac{7}{2\pi} \) Hz. The differential equation of S.H.M. is

• A. \( \frac{d^2x}{dt^2} + 14x = 0 \)
• B. \( \frac{d^2x}{dt^2} + 64x = 0 \)
• C. \( \frac{d^2x}{dt^2} + 49x = 0 \)
• D. \( \frac{d^2x}{dt^2} + 25x = 0 \)

Hint:

In S.H.M., the differential equation is always of the form \( \frac{d^2x}{dt^2} + \omega^2 x = 0 \), where \( \omega \) is the angular frequency.

Answer 1

The Correct Option is C

Step 1: Understanding the formula.
The frequency \( f \) of simple harmonic motion is related to the angular frequency \( \omega \) by: \[ f = \frac{\omega}{2\pi} \] Given that the frequency is \( \frac{7}{2\pi} \), we can find the angular frequency \( \omega \) as: \[ \omega = 7 \, \text{rad/s} \] Step 2: Relating angular frequency to the equation of motion.
The equation of motion for S.H.M. is: \[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \] Substituting \( \omega = 7 \), we get: \[ \frac{d^2x}{dt^2} + 49x = 0 \] Step 3: Conclusion.
Thus, the differential equation for the given frequency is \( \frac{d^2x}{dt^2} + 49x = 0 \), corresponding to option (C).