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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.

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    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

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