A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period of oscillation is (a) π s. (b) π/2 s. (c) 2π s. (d) π/t s.

This is a multiple choice answer as classified in NCERT Exemplar (a) V=dy/dt=awcoswt Vmax=aw=30 Acceleration A= d x 2 d t 2 = - a w 2 s i n w t A=w2a=60 So after solving w =2rad/s 2 π T = 2 r a d / s T= π s e c

A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². The period of oscillation is

(a) π s.

(b) π/2 s.

(c) 2π s.

(d) π/t s.

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Payal Gupta | Contributor-Level 10

7 months ago

This is a multiple choice answer as classified in NCERT Exemplar

(a) V=dy/dt=awcoswt

V_max=aw=30

Acceleration A= $\frac{d x^{2}}{d t^{2}} = - a w^{2} s i n w t$

A=w²a=60

So after solving w =2rad/s

2 $\frac{π}{T} = 2 r a d / s$

T= $π s e c$

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