The displacement of a particle of mass 2 text | vedclass

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Question

(A) Given mass $m = 2 	ext{ g} = 2 	imes 10^{-3} 	ext{ kg}$.The displacement equation is $x = 8 \cos \left(50 t + \frac{\pi}{12}ight) 	ext{ m}$.

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(A) Given mass $m = 2 	ext{ g} = 2 	imes 10^{-3} 	ext{ kg}$.The displacement equation is $x = 8 \cos \left(50 t + \frac{\pi}{12}ight) 	ext{ m}$.Comparing this with the standard $SHM$ equation $x = A \cos(\omega t + \phi)$,we get amplitude $A = 8 	ext{ m}$ and angular frequency $\omega = 50 	ext{ rad/s}$.The maximum velocity of the particle is $v_{\max} = A \omega = 8 	imes 50 = 400 	ext{ m/s}$.The maximum kinetic energy is given by $(K.E.)_{\max} = \frac{1}{2} m v_{\max}^2$.Substituting the values: $(K.E.)_{\max} = \frac{1}{2} 	imes (2 	imes 10^{-3} 	ext{ kg}) 	imes (400 	ext{ m/s})^2$.$(K.E.)_{\max} = 10^{-3} 	imes 160000 = 160 	ext{ J}$.

The displacement of a particle of mass $2 \text{ g}$ executing simple harmonic motion is $x = 8 \cos \left(50 t + \frac{\pi}{12}\right) \text{ m}$,where $t$ is time in seconds. The maximum kinetic energy of the particle is (in $\text{ J}$)

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