MHT CET · Physics · Oscillations
A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is \(16 \times 10^{-3} \mathrm{~J}\). The equation of motion of this particle is (Initial phase of oscillation \(=0^{\circ}\) )
- A \(Y=0.2 \sin (4 t)\)
- B \(Y=0.2 \sin \left(\frac{t}{4}\right)\)
- C \(Y=0.2 \sin \left(\frac{t}{2}\right)\)
- D \(Y=0.2 \sin (2 t)\)
Answer & Solution
Correct Answer
(D) \(Y=0.2 \sin (2 t)\)
Step-by-step Solution
Detailed explanation
\(KE_{max} = \frac{1}{2} m A^2 \omega^2\) \(16 \times 10^{-3} = \frac{1}{2} (0.2) (0.2)^2 \omega^2\)
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