Question

The differential equation of a particle performing S.H.M. is \(\frac{d^{2}x}{dt^{2}}+64x=0\), where \(x\) is in meter. The period of oscillation of the particle is (in second)

• A. \(\frac{\pi }{3}\)
• B. \(\frac{\pi }{4}\)
• C. \(\frac{\pi }{5}\)
• D. \(\frac{\pi }{8}\)

Hint:

Compare with the standard equation d2x/dt2 + omega squared x = 0 to get omega, then use T = 2 pi / omega.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The standard SHM equation is \(\dfrac{d^{2}x}{dt^{2}}+\omega^{2}x=0\).

Step 2: Find omega
\(\omega^{2}=64\), so \(\omega=8\text{ rad/s}\).

Step 3: Calculation
\(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{8}=\dfrac{\pi}{4}\text{ s}\)

Step 4: Compare the options
This is the second option.

Final Answer:
The period of oscillation is pi/4 seconds. \[ \boxed{\text{(B) }\dfrac{\pi}{4}} \]