AP EAMCET · PHYSICS · Oscillations
A particle executing simple harmonic motion has a maximum speed of \(40 \mathrm{~ms}^{-1}\) and maximum acceleration of \(60 \mathrm{~ms}^{-2}\). The period of oscillation is
- A \(\frac{4 \pi}{3} \mathrm{~s}\)
- B \(\frac{\pi}{2} \mathrm{~s}\)
- C \(2 \pi \mathrm{s}\)
- D \(\frac{1}{\pi} \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(\frac{4 \pi}{3} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given, maximum speed of SHM is \(40 \mathrm{~ms}^{-1}\). \(v_{\max }=a \omega=40 \mathrm{~ms}^{-1}\)...(i) Maximum acceleration of SHM is \(60 \mathrm{~ms}^{-2}\). As, \[ a_{\max }=a \omega^2=60 \mathrm{~ms}^{-2}...(ii) \] Dividing Eq. (ii) by Eq. (i), we get…
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