MHT CET · Physics · Oscillations
The equation of motion of a particle performing linear S.H.M is \(x=5 \sin \left[4 t-\frac{\pi}{6}\right]\), where \(x\) is its displacement in \(\mathrm{cm}\). The velocity of the particle when its displacement is \(3 \mathrm{~cm}\), is
- A \(8 \mathrm{~cm} / \mathrm{s}\)
- B \(6 \mathrm{~cm} / \mathrm{s}\)
- C \(16 \mathrm{~cm} / \mathrm{s}\)
- D \(10 \mathrm{~cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(16 \mathrm{~cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Comparing with standard equation:
\(x=a \sin [\omega t+\phi]\)
The amplitude of the wave is \(a=5\) and the angular velocity of the wave is \(\omega=4\) The velocity of the wave can be obtained by taking derivative as:
\(\begin{aligned} & \frac{d x}{d t}=\omega a \cos [\omega t+\phi]=\omega \sqrt{a^2-x^2} \\ & \therefore v=\omega \sqrt{a^2-y^2}=4 \sqrt{(5)^2-(3)^2}=16 \mathrm{~cm} / \mathrm{s}\end{aligned}\)
\(x=a \sin [\omega t+\phi]\)
The amplitude of the wave is \(a=5\) and the angular velocity of the wave is \(\omega=4\) The velocity of the wave can be obtained by taking derivative as:
\(\begin{aligned} & \frac{d x}{d t}=\omega a \cos [\omega t+\phi]=\omega \sqrt{a^2-x^2} \\ & \therefore v=\omega \sqrt{a^2-y^2}=4 \sqrt{(5)^2-(3)^2}=16 \mathrm{~cm} / \mathrm{s}\end{aligned}\)
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