The displacement time graph of a particle executing S.H.M. is shown in Fig. 14.5. Which of the following statement is/are true?

F1 savita Others 15-7-24 D16

  1. The force is zero at \(t = \frac{3T}{4}\).
  2. The acceleration is maximum at \(t = \frac{4T}{4}\).
  3. The velocity is maximum at \(t = \frac{T}{4}\).
  4. The P.E. is equal to K.E. of oscillation at \(t = \frac{T}{2}\).

Answer (Detailed Solution Below)

Option :

Detailed Solution

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Concept:

Simple harmonic motion:
Simple harmonic motion (S.H.M) is a type of periodic oscillation where the restoring force is directly proportional to the displacement.
→The motion of a simple pendulum is a good example of simple harmonic motion in which the bob oscillates freely along with the lowest and highest position under the influence of gravity.

Explanation:

The maximum velocity of a particle executing S.H.M. occurs at the mean position i.e. x = 0.
Thus velocity is maximum at T/4 and 3T/4, not T/2. Hence, force is zero

Therefore option (1) is correct.
Acceleration of a particle executing S.H.M. is maximum at extreme positions, thus at 0, T/2, T.

Hence, option (2) is correct.
The force or acceleration of a particle executing S.H.M.  is zero at the mean position, i.e. at T/4, 3T/4.

Hence, option (3) is correct.
KE=TE when PE=0 which occurs at the mean position. T/2 is extreme not mean,

Hence, option (4) is incorrect.

Hence, options (1), (2), and (3) are the correct answer.

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