Equation of SHM: Practice Problem & Solution
If a simple harmonic oscillator has got a displacement of $0.02 \text{ m}$ and acceleration equal to $2 \text{ m/s}^2$ at any time, the angular frequency of the oscillator is equal to: (1992)
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:
Magnitude of acceleration in SHM is $|a| = \omega^2|x|$.\nSubstitute the values: $2 = \omega^2 \times 0.02$.\n$\omega^2 = 100 \implies \omega = 10 \text{ rad/s}$.
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