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?

(a) The force is zero at t =3T/4 .

(b) The acceleration is maximum at t = 4T/4.

(c) The velocity is maximum at t = T/4.

(d) The P.E. is equal to K.E. of oscillation at t = T/2 .

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) At t= 3T/4, the displacement of the particle is zero. Hence particle executing SHM will be at mean position i.e x=0 acceleration is zero and force is also zero.

At t= 4T/3, displacement is maximum i.e extreme position, so acceleration is maximum

At t = T/4 corresponds to mean position, so velocity will be maximum at this position.

At t= T/2 corresponds to extreme position so KE =0 and PE =maximum.

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