Spring System in Harmonic Motion. How do I find

[TrolliOlli]

Homework Statement

A .5kg mass is suspended from a spring. A student attaches a second mass, .24kg, to the bottom of the first mass using tape (neglect the mass of the tape and spring). The spring extends an aditional 2cm when the second mass is attached.

a) Find the period of the small oscillations for the combined masses

b) The tape can only stand a force of 4N, you gradually increase the amplitude until the .24kg mass falls off, what is the amplitude when this occurs?

Homework Equations

F = ma
F= k(dx)
T = 2π√(m/k)

The Attempt at a Solution

a) F = Ma = k(dx)
(9.81) (.24) = k(.02), k = 117.72
T = 2π√(m/k) thus T = .498s

b) Since gravity is pulling the .24kg mass down, the spring will snap when the total force between gravity and the spring is 4n. This occurs when the spring is pulling up with a force of F = 4-9.81*.24 = 1.65N. I tried doing 1.65N = kx and solving for x, but I'm not giving any of the answers they give. What exactly am I doing wrong?

 

[Simon Bridge]

(a) is fine - for long answers the marker would prefer more explicit working.
(b) you need to draw a free body diagram for each mass, and work out the tension in the tape that way.

 

[TrolliOlli]

Ahh, I think I see my problem. It's because I forgot to subtract the weight of the larger mass from the force that the spring is exerting on the system.

Although when I try doing:

4N - 9.81*.24 + 9.81*.5 I get that the spring will have a force of 6.55 N when the 4N of tension is on the tape. Because of this, wouldn't I be able to take 6.55/k to get x? Still not getting the right answer doing this.

 

[Simon Bridge]

When in trouble - do the physics step-by-step.
FBD - sum forces - eliminate a - solve for x.

(and when talking to someone else - symbols are easier to read than numbers)