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

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(A) The maximum speed $(v_{\max})$ of a particle in simple harmonic motion is given by the formula $v_{\max} = A\omega$,where $A$ is the amplitude (ma

Vedclass · Accepted Answer

$1.59 \, mm$

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(A) The maximum speed $(v_{\max})$ of a particle in simple harmonic motion is given by the formula $v_{\max} = A\omega$,where $A$ is the amplitude (maximum displacement) and $\omega$ is the angular frequency.Given that $\omega = \frac{2\pi}{T}$,we can write $v_{\max} = A 	imes \frac{2\pi}{T}$.Rearranging for $A$,we get $A = \frac{v_{\max} 	imes T}{2\pi}$.Substituting the given values: $v_{\max} = 1.00 	imes 10^3 \, m/s$ and $T = 1.00 	imes 10^{-5} \, s$.$A = \frac{(1.00 	imes 10^3) 	imes (1.00 	imes 10^{-5})}{2\pi} = \frac{1.00 	imes 10^{-2}}{2\pi} \, m$.$A = \frac{0.01}{6.283} \approx 0.00159 \, m = 1.59 \, mm$.

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

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