i.) The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statement 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.
Answers
Given,
The graph shows variation of displacement of a particle performing S.H.M. with time t.
from figure, it is clear that graph of displacement of particle is a sinusoidal wave equation i.e., y = Asin(ωt)
velocity of particle, v = dy/dt = ωAcos(ωt) = ω√{A² - y²}, means velocity of particle is maximum when y = 0 [at mean position]
particle is in mean position at time T/2 [ see figure]
so, velocity of particle is maximum at T/2. therefore option (c) is incorrect.
velocity of particle is zero at extreme position i.e, x = A.
so, kinetic energy is zero at time T/4. therefore option (D) is also incorrect
acceleration of particle, a = d²y/dt² = -ω²Asin(ωt) = -ω²y , means magnitude of acceleration of particle is maximum at extreme position and is zero at mean position.
at time T, particle is at mean position. therefore acceleration of particle is zero at time T. so option (A) is also incorrect.
since magnitude of acceleration is maximum at extreme position, force is maximum at the extreme position.
at time 3T/4, force is maximum. therefore option (B) is correct.
Answer:
The force is maximum at time 3T/4.