An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the…

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:
  1. 3.0
  2. 2.0
  3. 1.5
  4. 1.0

Solution

$v_{\max }=15 \mathrm{~cm} / \mathrm{s}=15 \times 10^{-2} \mathrm{~m} / \mathrm{s}$ $T=628 \mathrm{~ms}=628 \times 10^{-3} \mathrm{~s}$ $v_{\max }=A \omega \Rightarrow v_{\max }=A \times \frac{2 \pi}{T}$ $15 \times 10^{-2}=A \times \frac{2 \times 3.14}{628 \times 10^{-3}}$ $A=\frac{15 \times 10^{-2} \times 628 \times 10^{-3}}{2 \times 3.14}$ $=15 \times 10^{-3} \mathrm{~m}$ $=1.5 \mathrm{~cm}$

Asked in: AP EAMCET 2003

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