Solving the Verticle Wave Problem: Time to Reach Equilibrium for a 0.2 kg Load

[timtng]

A load of mass .2 kg is hanging from a light spring whose elastic constant is 20 N/m. The load is pulled down .1 m from its equilibrium position and released.

How long is required for the load to reach its equilibrium position?

This is what I did:
T=2π(sqrt(m/k))=.628 s

my friend said the answer should be a fourth of that value, which is .157 s.

Is my friend correct?

Please help
Thanks

 

[yxgao]

Your friend is correct. The period you calculated is for one complete oscillation (that is, if the mass went pass equilibrium position, reached a stopping point, turned around, went pass equilibrium position again, and arrived at where you started from). From the stretched position, to get to equilibium requires 1/4 of a complete oscillation.

 

[M98Ranger]

No, your friend is not correct. Your calculation of 0.628 seconds is the correct answer. The equation T=2π√(m/k) is the formula for the period of a simple harmonic motion, which is the time it takes for the object to complete one full oscillation. In this case, the load will reach its equilibrium position after one full oscillation, so the time required for it to reach equilibrium is equal to the period. Therefore, your calculation of 0.628 seconds is the correct answer.