A particle of mass 0.1 kg is executing simpl | vedclass

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Question

(A) The mass of the particle is $m = 0.1 \ kg$. Amplitude of the particle performing $	ext{SHM}$ is $A = 0.1 \ m$. The initial phase of the particle

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The mass of the particle is $m = 0.1 \ kg$. Amplitude of the particle performing $	ext{SHM}$ is $A = 0.1 \ m$. The initial phase of the particle is $\phi = 45^{\circ} = \pi/4$. The general equation of a particle performing $	ext{SHM}$ is $x(t) = A \sin(\omega t + \phi)$. When the particle passes through the mean position,its kinetic energy is maximum,given by: $KE_{max} = \frac{1}{2} m \omega^2 A^2 = 8 	imes 10^{-3} \ J$. Substituting the values: $\frac{1}{2} 	imes 0.1 	imes \omega^2 	imes (0.1)^2 = 8 	imes 10^{-3}$. $0.05 	imes \omega^2 	imes 0.01 = 8 	imes 10^{-3}$. $0.0005 	imes \omega^2 = 0.008$. $\omega^2 = \frac{0.008}{0.0005} = 16$. $\omega = 4 \ rad/s$. Substituting $\omega$,$A$,and $\phi$ into the general equation: $x(t) = 0.1 \sin(4t + \pi/4)$.

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