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(C) The maximum speed of an object in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is t

Vedclass · Accepted Answer

$1.5$

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(C) The maximum speed of an object in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.We know that $\omega = \frac{2\pi}{T}$,where $T$ is the time period.Substituting this into the formula,we get $v_{\max} = a \left( \frac{2\pi}{T} ight)$.Rearranging for amplitude $a$,we have $a = \frac{v_{\max} 	imes T}{2\pi}$.Given values are $v_{\max} = 15 \, cm/s$ and $T = 628 \, ms = 628 	imes 10^{-3} \, s = 0.628 \, s$.Using $\pi \approx 3.14$,we calculate $a = \frac{15 	imes 0.628}{2 	imes 3.14} = \frac{15 	imes 0.628}{6.28} = \frac{15 	imes 0.628}{10 	imes 0.628} = \frac{15}{10} = 1.5 \, cm$.

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 \, cm/s$ and the period is $628 \, ms$. The amplitude of the motion in centimeters is

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