Spring/Mass System: Finding Displacement Equations

[foxjwill]

Homework Statement

Two blocks of equal mass and negligible volume connected by an ideal spring are set on a horizontal, frictionless surface. Block 1 is flush against a wall. Find an equation for the displacement of each block, in terms of amplitude, angular frequency, and time, if Block 2 has compressed the spring such that both blocks are touching.

Homework Equations

The Attempt at a Solution

I've attached all my work -- it would've taken way to much time to $$\TeX \mbox{ify}$$ all of it. I was wondering if someone could check to make sure my answer's correct.

 

[AvroArrow]

Thanks!Answer: The displacement of Block 1 is given by$x_1(t) = \frac{A}{2}\cos(\omega t)$The displacement of Block 2 is given by$x_2(t) = \frac{A}{2} \cos(\omega t + \pi)$

 

[Moetasim]

I would first note that this problem can be solved using the principles of simple harmonic motion. The displacement equations for each block can be derived by considering the forces acting on each block.

For Block 1, the only force acting on it is the force exerted by the spring. This force can be described by Hooke's Law, F = -kx, where k is the spring constant and x is the displacement from the equilibrium position. Since Block 1 is flush against the wall, its displacement is 0. Therefore, the equation for Block 1's displacement can be written as x1 = A1sin(ωt), where A1 is the amplitude and ω is the angular frequency.

For Block 2, there are two forces acting on it: the force exerted by the spring and the force exerted by Block 1. The force exerted by the spring can again be described by Hooke's Law, F = -kx. The force exerted by Block 1 can be calculated using Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Therefore, the force exerted by Block 1 on Block 2 is equal in magnitude but opposite in direction to the force exerted by Block 2 on Block 1. This force can be described by F = kx, where x is the displacement of Block 1.

Combining these two forces, we can write the equation for Block 2's displacement as x2 = A2sin(ωt) + A1sin(ωt), where A2 is the amplitude of Block 2's displacement.

To solve for the amplitudes, we can use the fact that at the point where both blocks are touching, the total displacement of the spring is equal to 0. This means that A1 + A2 = 0, or A2 = -A1. Substituting this into the equation for Block 2's displacement, we get x2 = A1(sin(ωt) - sin(ωt)) = 0. This makes sense, as at this point, both blocks are not moving and their displacement is 0.

In conclusion, the equations for the displacement of each block are x1 = A1sin(ωt) and x2 = -A1sin(ωt). These equations show that the displacement of each block is sinusoidal and that they