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KCET · Physics · Oscillations

The displacement of a particle executing SHM is given by \(x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]\), where \(x\) is in metre and \(t\) is in seconds. The amplitude and maximum speed of the particles is

  1. A \(3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}\)
  2. B \(3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}\)
  3. C \(3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}\)
  4. D \(3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}\)

Step-by-step Solution

Detailed explanation

Given, displacement equation of particle executing SHM,
\(X=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]\)
Comparing with displacement equation of SHM as \(y=A \sin (\omega t+\phi)\), we get
Amplitude, \(A=3 \mathrm{~m}\)
Angular velocity, \(\omega=2 \pi \mathrm{rad} / \mathrm{s}\)
\(\therefore\) Maximum speed,
\(v_{\max }=\omega A=2 \pi \times 3=6 \pi \mathrm{m} / \mathrm{s}\)