A 1.00 times 10^{-20} ,kg particle is vibrati | vedclass

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(A) Given:Mass $m = 1.00 	imes 10^{-20} \,kg$Period $T = 1.00 	imes 10^{-5} \,s$Maximum speed $v_{max} = A \omega = 1.00 	imes 10^3 \,m/s$Angular f

Vedclass · Accepted Answer

$1.59$

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(A) Given:Mass $m = 1.00 	imes 10^{-20} \,kg$Period $T = 1.00 	imes 10^{-5} \,s$Maximum speed $v_{max} = A \omega = 1.00 	imes 10^3 \,m/s$Angular frequency $\omega$ is given by:$\omega = \frac{2\pi}{T} = \frac{2\pi}{1.00 	imes 10^{-5}} = 2\pi 	imes 10^5 \,rad/s$Using the formula for maximum speed:$v_{max} = A \omega$$1.00 	imes 10^3 = A 	imes (2\pi 	imes 10^5)$Solving for amplitude $A$:$A = \frac{1.00 	imes 10^3}{2\pi 	imes 10^5} = \frac{1}{2\pi} 	imes 10^{-2} \,m$$A \approx 0.15915 	imes 10^{-2} \,m = 1.5915 	imes 10^{-3} \,m$Since $1 \,m = 1000 \,mm$:$A = 1.5915 	imes 10^{-3} 	imes 10^3 \,mm = 1.5915 \,mm$Rounding to two decimal places,$A = 1.59 \,mm$.

$A$ $1.00 \times 10^{-20} \,kg$ particle is vibrating under simple harmonic motion with a period of $1.00 \times 10^{-5} \,s$ and with a maximum speed of $1.00 \times 10^3 \,m/s$. The maximum displacement of the particle from the mean position is .......... $mm$.

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