KCET · Physics · Oscillations
A particle executing a simple harmonic motion has a period of \(6 \mathrm{~s}\). The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is
- A \(\frac{3}{2} \mathrm{~s}\)
- B \(\frac{1}{2} \mathrm{~s}\)
- C \(\frac{3}{4} \mathrm{~s}\)
- D \(\frac{1}{4} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Equation for simple harmonic motion
\(y =a \sin \left(\frac{2 \pi}{T}\right) t \)
\( \frac{a}{2} =a \sin \left(\frac{2 \pi}{T}\right) t \)
\( \frac{2 \pi}{T} t =\frac{\pi}{6} \)
\( t =\frac{T}{12}=\frac{6}{12}=\frac{1}{2} \mathrm{~s}\)
\(y =a \sin \left(\frac{2 \pi}{T}\right) t \)
\( \frac{a}{2} =a \sin \left(\frac{2 \pi}{T}\right) t \)
\( \frac{2 \pi}{T} t =\frac{\pi}{6} \)
\( t =\frac{T}{12}=\frac{6}{12}=\frac{1}{2} \mathrm{~s}\)
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