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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/s^2. The period of oscillation is (a) π s.

A particle executing S.H.M. has a maximum speed of 30 \frac{cm}{s} and a maximum acceleration of 60 \frac{cm}{s^{2}}. The period of oscillation is
(a) \pi s.
(b) \left ( \frac{\pi}{2} \right ) s.
(c) 2\pi s.
(d) \left (\frac{\pi}{t} \right ) s.

Answers (1)

Answer: The answer is the option (a) \pi s

Explanation: Let us consider the equation of a SHM

Thus,y = a \sin \omega t

v = \frac{dy}{dt}

      = a\omega \cos \omega t    ……. (i)

a = \frac{dv}{dt}

      = -a\omega^ 2 \sin \omega t   ….. (ii)

V_{max} = 30 \frac{cm}{s}     …. (given)

From (i),

V_{max} = a \omega

Thus, a \omega=30     ….. (iii)

From (ii),

Now, a_{max} = a\omega ^{2} = 60  …….. (iv)

Thus, 60 = \omega \times 30

Therefore, \omega = 2 \frac{rad}{s}

Now, \frac{2 \pi }{T} = 2

Thus, T = \pi.

Hence opt (a).

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