Q129 / 160 · AP EAMCET 2003 · Physics

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:

  • A
    3.0
  • B
    2.0
  • Correct answer
    C
    1.5
  • D
    1.0

Solution

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

Hence, the correct answer is option: 'C'.

Relevant Questions