Another Harmonic Motion Problem

[elsternj]

Homework Statement

On a horizontal, frictionless table, an open-topped 5.50kg box is attached to an ideal horizontal spring having force constant 365 N/m . Inside the box is a 3.44-kg stone. The system is oscillating with an amplitude of 7.50 . When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box.
Find the period of the resulting motion of the box.

Homework Equations

T = 2pi\sqrt{m/C}

The Attempt at a Solution

okay well for the first half of the period the mass is 3.44+5.5 = 8.94

.5T = 2pi\sqrt{8.94/365} = .983

then the stone is thrown out

.5T = 2pi\sqrt{5.5/365} = .771

.771 + .983 = 1.75s

wrong. I am new to this harmonic motion so I could be way off. any guidance?

 

[eshaw]

Since the period in a harmonic oscillator only depends on the mass and the spring constant, as soon as the mass changes the period will change almost instantly and won't depend on what the period was before. So, I guess the answer should be .771 based on your work. The period was .983 and now it is .771.

 

[elsternj]

awesome thanks! that makes sense.

I am now having trouble finding the amplitude

if x = Acos(\sqrt{k/m}t)

x = Acos(8.15*.771)

however i still have 2 unknowns here. x and A. i can't see to find another formula in my book that will help with this. Everywhere I find A I also find x.

 

[eshaw]

Finding the amplitude is a bit harder, since it does depend on the starting conditions. Do you know how to solve differential equations and apply starting conditions? I'm going to assume not. So, x is a function of time for the position of the box, the t isn't the period in that equation, it is the time. The derivative of that equation will be the velocity. What you should find is the speed at the point when the stone is taken out. Using the concept of total energy of the system E=K+U, it turn's out that the energy of the harmonic oscillator E is also equal to (kA^2)/2 (you should find this by plugging in x and it's derivative into the equations for kinetic energy and potential energy). At the point when the stone is taken out there is no potential energy so you can find the amplitude in this way: (kA^2)/2=(mv^2)/2. I probably gave too much information, but you should figure out how to show the work for the steps that I left out.