Question

a body is executing simple harmonic motion with an angular frequency of 2 Rad/sec. The velocity of the body at 20mm displacement, when the amplitude of the motion is 60 mm is?
can u help me with solution?

Answer 1

The equation of motion:

$y=A \sin \omega t\\</p><p>A=60 \times 10^{-3}m\\</p><p>\omega=2\\</p><p>\frac{dy}{dt}=v=A\omega \cos \omega t\\</p><p>v=(6\times 10^{-2}) \times 2 \times \sqrt{2}/3\\</p><p>v=4\sqrt{2} \times 10^{-2}m/sec$

Answer 2

Answer:

v=\omega \sqrt{({{a}^{2}}-{{y}^{2}})}=2\sqrt{{{60}^{2}}-{{20}^{2}}}=113\,mm/

Explanation: