TS EAMCET · Physics · Oscillations
A particle performs simple harmonic motion with a time period of \(16 \mathrm{~s}\). At a time \(t=2 \mathrm{~s}\), the particle passes through the origin and at \(\mathrm{t}=4 \mathrm{~s}\) its velocity is \(4 \mathrm{~m} / \mathrm{s}\). The amplitude of the motion is
- A \(\frac{32 \pi}{\sqrt{2}}\)
- B \(\frac{32 \sqrt{2}}{\pi}\)
- C \(32 \pi\)
- D 32
Answer & Solution
Correct Answer
(B) \(\frac{32 \sqrt{2}}{\pi}\)
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