A) 0.08 m/s
B) 0.05 m/s
C) 0.04 m/s
D) 0.01 m/s
Correct Answer: A
Solution :
When the point is at a distance from the mean position its, velocity is given by \[v=\omega \,\sqrt{{{a}^{2}}-{{y}^{2}}}\] Given, \[T=\frac{2\pi }{\omega }=\pi \] \[\therefore \] \[\omega ={{25}^{-1}}\] and \[v=a\omega \] \[0.1=a\times 2\] \[a=\frac{0.1}{2}=0.05\,m\] y = 0.05 m \[\therefore \] Velocity, \[v=2\sqrt{{{(0.05)}^{2}}-{{(0.03)}^{2}}}\] = 0.08 m/s
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