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(B) The given equation is $x = \sin \pi (t + 1/3) \, m$.Expanding the equation,we get $x = \sin (\pi t + \pi/3) \, m$.The velocity $v$ is the derivati

Vedclass · Accepted Answer

$157$

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(B) The given equation is $x = \sin \pi (t + 1/3) \, m$.Expanding the equation,we get $x = \sin (\pi t + \pi/3) \, m$.The velocity $v$ is the derivative of displacement $x$ with respect to time $t$:$v = \frac{dx}{dt} = \frac{d}{dt} [\sin (\pi t + \pi/3)] = \pi \cos (\pi t + \pi/3) \, m/s$.At $t = 1 \, s$:$v = \pi \cos (\pi(1) + \pi/3) = \pi \cos (4\pi/3) \, m/s$.Since $\cos (4\pi/3) = \cos (\pi + \pi/3) = -\cos (\pi/3) = -1/2$:$v = \pi 	imes (-1/2) = -\pi/2 \, m/s$.The speed is the magnitude of velocity:$|v| = |-\pi/2| = \pi/2 \, m/s$.Converting to $cm/s$ $(1 \, m = 100 \, cm)$:$|v| = (\pi/2) 	imes 100 = 50\pi \, cm/s$.Given $\pi = 3.14$:$|v| = 50 	imes 3.14 = 157 \, cm/s$.

The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

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