A point particle of mass 0.1, kg is executing | vedclass

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(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} = a\omega \cos(\o

Vedclass · Accepted Answer

$y = 0.1 \sin (\pm 4t + \pi/4)$

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 $v = \frac{dy}{dt} = a\omega \cos(\omega t + \phi)$.The velocity is maximum at the mean position,where $v_{max} = a\omega$.The kinetic energy at the mean position is $K.E._{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (a\omega)^2$.Given $m = 0.1 \, kg$,$a = 0.1 \, m$,and $K.E._{max} = 8 	imes 10^{-3} \, J$.Substituting these values: $\frac{1}{2} 	imes 0.1 	imes \omega^2 	imes (0.1)^2 = 8 	imes 10^{-3}$.$0.05 	imes 0.01 	imes \omega^2 = 8 	imes 10^{-3} \implies 0.0005 	imes \omega^2 = 0.008$.$\omega^2 = \frac{0.008}{0.0005} = 16 \implies \omega = 4 \, rad/s$.The initial phase $\phi = 45^o = \pi/4 \, rad$.Thus,the equation of motion is $y = 0.1 \sin(\pm 4t + \pi/4)$.

$A$ 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 $45^o$.

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