Simple Harmonic Motion: Calculating Speeds of Blocks with a Compressed Spring

[pezola]

Homework Statement

A 15.8 kg block and a 31.6 kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant = 1868 N/m). The spring is compressed by 0.152 m from its unstrained length and is not attached permanently to either block. With what speed does each block move away when the mechanism keeping the spring squeezed is released and the spring falls away?

The Attempt at a Solution

This is a sample problem for the final exam. Please explain the process, I'm not just looking for a numerical answer.

 

[Doc Al]

Show what you've done so far and point out where you are stuck.

Hint: What's conserved?

(This isn't a SHM problem.)

 

[Moetasim]

As a scientist, it is important to approach problems like this using the principles of physics and mathematics. In this case, we can use the concept of simple harmonic motion to calculate the speeds of the two blocks when the spring is released.

First, we can start by drawing a free body diagram for each block. We know that the forces acting on the blocks are the force of gravity (mg) and the force of the spring (F = kx), where k is the spring constant and x is the displacement of the spring from its unstrained length.

Next, we can apply Newton's second law (ΣF = ma) to each block. Since the blocks are initially at rest, we can set the net force equal to zero. This allows us to solve for the acceleration of each block.

For the 15.8 kg block, we have:

ΣF = ma
0 = mg - kx
a = g - kx/m

For the 31.6 kg block, we have:

ΣF = ma
0 = mg + kx
a = -g + kx/m

Now, using the equation for simple harmonic motion (a = -ω^2x), we can solve for the angular frequency (ω) of the oscillation of the blocks. This can be done by equating the expressions for acceleration in terms of ω:

g - kx/m = -ω^2x
g + kx/m = -ω^2x

Solving for ω, we get:

ω = √(k/m)

Now, we can use this value of ω to calculate the speeds of the blocks using the equation v = ωx. For the 15.8 kg block, we have:

v = √(k/m)x
v = √(1868/15.8)(0.152)
v = 1.49 m/s

For the 31.6 kg block, we have:

v = √(k/m)x
v = √(1868/31.6)(0.152)
v = 0.94 m/s

Therefore, when the mechanism keeping the spring squeezed is released, the 15.8 kg block will move away with a speed of 1.49 m/s and the 31.6 kg block will move away with a speed of 0.94 m/s. It is important to note