Question

The displacement and acceleration of a particle oscillating simple harmonically are respectively 3.0 m and \(48.0 \, ms^{-2}\). The angular frequency of the particle is (in \(s^{-1}\)).

• A. 6
• B. 3
• C. 5
• D. 2
• E. 4

Hint:

Remember $a = -\omega^2 x$. The negative sign indicates direction; for calculations, use absolute values.

Answer 1

Answer: (E)

Step 1: Concept
In Simple Harmonic Motion, the magnitude of acceleration is related to displacement by $a = \omega^2 x$.

Step 2: Meaning
Angular frequency ($\omega$) can be found by taking the square root of the ratio of acceleration to displacement.

Step 3: Analysis
$\omega^2 = a/x = 48.0 / 3.0 = 16$.
$\omega = \sqrt{16} = 4~s^{-1}$.

Step 4: Conclusion
Hence, the angular frequency is 4.
Final Answer: (E)