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
- A \(3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}\)
- B \(3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}\)
- C \(3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}\)
- D \(3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}\)
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,
\(v_{\max }=\omega A=2 \pi \times 3=6 \pi \mathrm{m} / \mathrm{s}\)
\(X=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]\)
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,
\(v_{\max }=\omega A=2 \pi \times 3=6 \pi \mathrm{m} / \mathrm{s}\)
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