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Physics Question on simple harmonic motion

The equation of a particle executing simple harmonic motion is given by x=sin  π(t+13)mx=sin\;π\bigg(t+\frac{1}{3}\bigg)m. At t=1st = 1s, the speed of the particle will be

A

0  cm  s−10\;cm\;s^{-1}

B

157  cm  s−1157\;cm\;s^{-1}

C

272  cm  s−1272\;cm\;s^{-1}

D

314  cm  s−1314\;cm\;s^{-1}

Answer

157  cm  s−1157\;cm\;s^{-1}

Explanation

Solution

Speed of the Particle

x=sin(πt+π3)mx=sin\bigg(πt+\frac{π}{3}\bigg)m

⇒dxdt=πcos(πt+π3)⇒\frac{dx}{dt}=πcos\bigg(πt+\frac{π}{3}\bigg)

=πcos(π+π3)=πcos\bigg(π+\frac{π}{3}\bigg)

at t=1st = 1s

=–π2  m/s–\frac{π}{2}\;m/s

or ∣dxdt∣=157  cm/s|\frac{dx}{dt}| = 157 \;cm/s