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

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



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

Answers (1)

The answer is the option (a) The force is zero at t = \left (\frac{3T}{4} \right ), (b) The acceleration is maximum at t = \left (\frac{4T}{4} \right ), and (c) The velocity is maximum at t = \left (\frac{T}{4} \right )

Explanation: (a) The particle is at its mean position at t = \left (\frac{3T}{4} \right ), so the force acting on it zero, but due to inertia of mass the motion continues

a = 0 thus, F = 0.

(b) Particles velocity changes increasing to decrease so maximum at the change in velocity, at t = \left (\frac{4T}{4} \right ). Thus, acceleration is maximum here.

(c) The velocity is maximum at its mean position at t = \left (\frac{T}{4} \right ) as there is no retarding force on it.

(d) K.E. = 0 at t = \frac{T}{2} = \frac{2T}{4}.

Thus, P.E is not equal to kinetic energy.

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