AP EAMCET · PHYSICS · Oscillations
The maximum velocity of a particle performing simple harmonic motion is \(6.28 \mathrm{~cm} \mathrm{~s}^{-1}\). If the length of its path is \(8 \mathrm{~cm}\), then what is its period?
- A \(2 \mathrm{~s}\)
- B \(4 \mathrm{~s}\)
- C \(3 \mathrm{~s}\)
- D \(1 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(4 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
In SHM, \[ v_{\max }=6.28 \mathrm{~cm} \mathrm{~s}^{-1} \] Length of path \(=8 \mathrm{~cm}\) \(\therefore\) Amplitude of particle, \(a=\frac{\text { Length of path }}{2}\) \[ =\frac{8}{2}=4 \mathrm{~cm} \]…
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