A point particle of mass 0.1 kg is executing | vedclass

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Question

(A) The displacement of a particle in $S.H.M.$ is given by: $y = a \sin (\omega t + \phi)$.The velocity is given by $v = \frac{dy}{dt} = \omega a \cos

Vedclass · Accepted Answer

$y=0.1 \sin \left(\pm 4 t+\frac{\pi}{4}\right)$

Vedclass · Answer

(A) The displacement of a particle in $S.H.M.$ is given by: $y = a \sin (\omega t + \phi)$.The velocity is given by $v = \frac{dy}{dt} = \omega a \cos (\omega t + \phi)$.The velocity is maximum at the mean position,where $v_{max} = \omega a$.The kinetic energy at the mean position is $K.E._{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m \omega^2 a^2$.Given $m = 0.1 \ kg$,$a = 0.1 \ m$,and $K.E._{max} = 8 	imes 10^{-3} \ J$,we have:$8 	imes 10^{-3} = \frac{1}{2} 	imes 0.1 	imes \omega^2 	imes (0.1)^2$.$8 	imes 10^{-3} = 0.5 	imes 0.1 	imes \omega^2 	imes 0.01$.$8 	imes 10^{-3} = 0.0005 	imes \omega^2$.$\omega^2 = \frac{8 	imes 10^{-3}}{5 	imes 10^{-4}} = 16$.$\omega = \pm 4 \ rad/s$.Given the initial phase $\phi = 45^{\circ} = \frac{\pi}{4} \ rad$,the equation of motion is $y = 0.1 \sin (\pm 4t + \frac{\pi}{4})$.

$A$ point particle of mass $0.1 \ kg$ is executing $S.H.M.$ with an amplitude of $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^{\circ}$.

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