AP EAMCET · PHYSICS · Oscillations
A particle of mass \(0.4 \mathrm{~kg}\) executes simple harmonic motion of amplitude \(0.4 \mathrm{~m}\). When it passes through the mean position, its kinetic energy is \(256 \times 10^{-3} \mathrm{~J}\). If the initial phase of the oscillation is \(\pi / 4\), then the equation of its motion is
- A \(x=0.4 \sin \left((0.4) t+\frac{\pi}{4}\right)\)
- B \(x=02 \sin \left(2 \sqrt{2}+\left(\frac{\pi}{4}\right) t\right)\)
- C \(x=0.8 \sin \left((2 \sqrt{2}) t+\frac{\pi}{2}\right)\)
- D \(x=0.4 \sin \left((2 \sqrt{2}) t+\frac{\pi}{4}\right)\)
Answer & Solution
Correct Answer
(D) \(x=0.4 \sin \left((2 \sqrt{2}) t+\frac{\pi}{4}\right)\)
Step-by-step Solution
Detailed explanation
Given, mass of the particle, \(m=0.4 \mathrm{~kg}\) Amplitude, \(A=0.4 \mathrm{~m}\) Initial phase, \(\phi=\frac{\pi}{4}\) Kinetic energy at mean position \(\mathrm{KE}\) As,…
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