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An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 \mathrm{~cm} / \mathrm{s}$ and the period is 628 milli-seconds. The amplitude of the motion in cm is:
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The correct answer is:
1.5
$v_{\max }=15 \mathrm{~cm} / \mathrm{s}=15 \times 10^{-2} \mathrm{~m} / \mathrm{s}$
$T=628 \mathrm{~ms}=628 \times 10^{-3} \mathrm{~s}$
$v_{\max }=A \omega \Rightarrow v_{\max }=A \times \frac{2 \pi}{T}$
$15 \times 10^{-2}=A \times \frac{2 \times 3.14}{628 \times 10^{-3}}$
$A=\frac{15 \times 10^{-2} \times 628 \times 10^{-3}}{2 \times 3.14}$
$=15 \times 10^{-3} \mathrm{~m}$
$=1.5 \mathrm{~cm}$
$T=628 \mathrm{~ms}=628 \times 10^{-3} \mathrm{~s}$
$v_{\max }=A \omega \Rightarrow v_{\max }=A \times \frac{2 \pi}{T}$
$15 \times 10^{-2}=A \times \frac{2 \times 3.14}{628 \times 10^{-3}}$
$A=\frac{15 \times 10^{-2} \times 628 \times 10^{-3}}{2 \times 3.14}$
$=15 \times 10^{-3} \mathrm{~m}$
$=1.5 \mathrm{~cm}$
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