Solving for Period of Force Pulled Downward and Released - 27N Weight

[vpv]

A body of weight 27 N hangs on a long spring of such stiffness that an extra force of 9N stretches the spring 0.05 m. If the body is pulled downward and released what is its period?

I know how to do this question, except I am not understanding the concept entirely. First I would calculate k, the spring constant.
k = F / x = 36 / 0.05
I don't understand exactly why the answer book is using only 9 N as the force. I would think that if you have 9 N alone on the spring, then the spring would not stretch but if you 9 N in addition to the 27 N already there, then can the spring stretch. So wouldn't the total force on the spring required to stretch it by 0.05 m be 36 N and not 9 N? Thank you.

 

[jbsteele]

Yes, you are correct. The total force required to stretch the spring by 0.05 m is 36 N. In this case, the force of 9 N is the additional force needed to stretch the spring beyond its original length due to the 27 N weight hanging from it. The spring constant for this problem can then be calculated as k = F / x = 36 / 0.05. Once the spring constant is known, the period of the oscillation can be calculated using the equation T = 2π√(m/k).