Simple harmonic motion spring problem

[Dylan.Wallett]

Homework Statement

The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass*m1*and*m2*attached to them. Each spring has a spring constant of*120*N/m. The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at*x*= 0 m if the following masses are used: m1 = 3.0 kg m2 = 2.7 kg

Homework Equations

w = squ root k/m - frequency of vibration
w = 2 pi/T
E=1/2mv^2 + 1/2kx^2

The Attempt at a Solution

I am pretty stumped with this one. I feel we need to know the distance that the two particles were pulled to and then released.
I started with
m1 - w= squ root k/m
= squ root 120/3.0
= 2.0 rad/s

m2 - w= squ root k/m
= squ root 120/2.7
= 2.1 rad/s

Thanks

 

[gneill]

I can't find your drawing. Are you sure that it was properly attached?

 

[Dylan.Wallett]

Sorry - not sure what happened. I have attached it again. Hope it comes through this time

 

[gneill]

Much better.

The frequencies and periods of the two spring-mass systems are independent of their amplitudes, depending only upon the masses and spring constants. If you let ω1 and ω2 be their angular frequencies respectively, then you should be able to write an equation for the position of each with respect to time as sinusoidal functions. Since you're only concerned about where these functions are zero, you can just set their amplitudes to unity.

 

[Dylan.Wallett]

I see now that the one particle will be going slower than the other and therefore if released together at the same time the amplitudes will be in unity and will create two sinusoidal functions that will meet at zero.
So I will make w1 and w2 = f1 and f2. Find the periods (which should be equal) and then find out the time at X=0. I am going to draw it out and try and find out if there are mistakes I have written here.

Thanks for putting me on the right track.