MCQ · Q5
Q.The graph shows the variation of displacement of a particle performing S.H.M. with time t, starting from the mean position (x = 0 at t = 0) and rising first towards the positive extreme, i.e. x = A sin(2\pi t/T), so that the particle is at the mean position at t = 0, T/2, T, ... and at an extreme position at t = T/4, 3T/4, ... . Which of the following statements is correct from the graph? (A) The acceleration is maximum at time T. (B) The force is maximum at time 3T/4. (C) The velocity is zero at time T/2. (D) The kinetic energy is equal to total energy at time T/4.
Maharashtra MsbshseNCERTSubjectiveImportance★★★★★est
27% · 15/55 Questions
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Concept understanding — Displacement, Velocity and Acceleration in SHM
For a particle in linear S.H.M., displacement x, velocity v and acceleration a are all determined at every instant by a single phase angle (omega t + phi): x = A sin(omega t + phi), v = Aomegacos(omega t + phi), a = -Aomega^2sin(omega t + phi) = -omega^2 x. Two starting conditions come up constantly: starting from the mean position (x = 0 at t = 0) gives phi = 0 and x …
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