Equation of SHM: Practice Problem & Solution
The equation of SHM of a particle is given as 2d²x/dt² + 32x = 0, where x is the displacement from the mean position. then time period of its oscillation (in seconds) is
Solution Explained:
To solve this problem, we apply the core principles of Equation of SHM. Understanding the underlying formula is key to arriving at the correct answer below:
Given the SHM equation:
\[
2 \frac{d^2x}{dt^2} + 32x = 0
\]
We can rewrite it as:
\[
\frac{d^2x}{dt^2} + 16x = 0
\]
This equation is of the form:
\[
\frac{d^2x}{dt^2} + \omega^2 x = 0
\]
where \( \omega^2 = 16 \).
Solution:
1. Angular Frequency \( \omega \):
\[
\omega = \sqrt{16} = 4 \, \text{rad/s}
\]
2. Time Period \( T \):
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s}
\]
Answer:
The time period of the oscillation is \( \frac{\pi}{2} \) seconds.
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