A 0.020 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 180 N/m. The block is pulled from its equilibrium position at x=0 m to a displacement x=+0.080 m and its released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t=0.50 s is closest to:

a.) -3m/s b.) -7 m/s c.) zero d.) 3 m/s e.) 7 m/s

Question

A 0.020 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 180 N/m. The block is pulled from its equilibrium position at x=0 m to a displacement x=+0.080 m and its released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t=0.50 s is closest to:

a.) -3m/s  b.) -7 m/s  c.) zero d.) 3 m/s e.) 7 m/s

anonymous · Answer

In a spring mass simple harmonic system, the angular velocity is given by $\frac{m}{k}$, where the velocity is obtained from differentiating $$x=r cos \omega  t$$
so, $v=-r \omega { sin{ \omega t}}$
Substitute all the given values to find the velocity. Have you attempted it?

anonymous · Answer

I think angular velocity is $$\omega = \sqrt{\frac{ k }{ m }}$$
But apart from that solution is correct (worth mentioning that r is the initial displacement).

anonymous · Answer

I don't understand how to do this......

anonymous · Answer

nevermind. The correct formula to use is Velocity=-wAsinwt, the answer is closest to 3m/s. Thanks

anonymous · Answer

Right sorry about that. thanks @furnessj