An object is attached to the bottom of a ligh | vedclass

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(C) Given: Maximum speed $v_{\max} = 15 ~cm/s$ and period $T = 628 ~ms = 0.628 ~s$. In simple harmonic motion,the maximum speed is given by $v_{\max}

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$1.5$

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(C) Given: Maximum speed $v_{\max} = 15 ~cm/s$ and period $T = 628 ~ms = 0.628 ~s$. In simple harmonic motion,the maximum speed is given by $v_{\max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency. We know that $\omega = \frac{2\pi}{T}$. Substituting the value of $\omega$ into the formula: $v_{\max} = A 	imes \frac{2\pi}{T}$. Rearranging for amplitude $A$: $A = \frac{v_{\max} 	imes T}{2\pi}$. Substituting the given values: $A = \frac{15 ~cm/s 	imes 0.628 ~s}{2 	imes 3.14}$. $A = \frac{15 	imes 0.628}{6.28} ~cm$. $A = \frac{15 	imes 0.628}{10 	imes 0.628} ~cm = 1.5 ~cm$.

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 $cm$ is:

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