Two masses connected by a spring

In summary, two masses of 2kg and 4kg are placed on opposite ends of a compressed spring with a constant of 3N/m. At the other end of the plane is a second spring with a constant of 12N/m. The velocities of the masses are found to be -0.016m/s and 0.008m/s respectively, and the maximum compression of the second spring is 0.0046m. The given values in the textbook are incorrect.
  • #1
grusini
7
0

Homework Statement


Two masses ##m_1=2kg##, ##m_2=2m_1## are placed on the opposite edges of a spring of constant ##k_1=3N/m##, compressed of a length ##x_1=1.73cm##. The system is located on a smooth plane. At the right end of the plane there is a second spring of constant ##k_2=12N/m##. Once the first spring is at rest the masses are free to move (they are not fixed to the spring). So the situation is as follows:

_________________M1 spring1 M2___________spring2 ||

a) Find the velocity ##v_1,v_2## of masses ##m_1,m_2## respectively, when the first spring is at rest.

b) Find the maximum compression ##\Delta x## of the second spring.

Homework Equations


Conservation of energy, conservation of momentum, potential energy of a spring


The Attempt at a Solution



a) The velocities must satisfy the system:
$$\begin{cases}\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}k_1x_1^2 \\
m_1v_1+m_2v_2=0\end{cases}$$
##\Rightarrow v_2=\sqrt{\frac{m_1k_1x_1^2}{m_2^2+m_1m_2}}=##
##\sqrt{\frac{2\cdot 3\cdot (0.0173)^2}{4^2+4\cdot 2}}=0.008 m/s##.
Then ##v_1=-\frac{m_2}{m_1}v_2=-2v_2=-0.016m/s##.

b) The value ##\Delta x## must satisfy:
##\frac{1}{2}m_2v_2^2=\frac{1}{2}k_2(\Delta x)^2 \Rightarrow \Delta x=\sqrt{\frac{m_2v_2^2}{k_2}}=\sqrt{\frac{4\cdot (0.008)^2}{12}}=0.0046m##

Values are wrong on my textbook..
 
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  • #2
What do you mean the values are wrong in your textbook? Do you mean your textbook gives you a solution that is different from your own? What is the given solution?

Your work seems to make sense to me. Perhaps you shouldn't be rounding the value you get for v_2 when you work through the whole problem, ie pretend 0.00865 m/s is entirely significant until the end (also, you should probably round that to 0.009 if you're going to round)
 
  • #3
I talked to my professor and he agrees that the values found in my textbooks are wrong (what I meant before was that my result differs from that of the textbook). Anyway, thank you Alucinator for the tip about the rounding!
 

1. How does the spring affect the motion of the two masses?

The spring acts as a restoring force, pulling the masses towards each other when they are stretched apart and pushing them away when they are compressed. This causes the masses to oscillate back and forth around their equilibrium position.

2. What factors affect the frequency of the oscillations?

The frequency of oscillations is affected by the mass of the masses, the stiffness of the spring, and the amplitude of the oscillations. As the mass or stiffness increases, the frequency decreases. As the amplitude increases, the frequency increases.

3. Can the spring ever reach its natural length during the oscillations?

No, the spring will always be stretched or compressed to some degree during the oscillations. However, as the amplitude decreases, the spring will come closer to its natural length.

4. How is energy conserved in this system?

In a frictionless environment, the total mechanical energy (kinetic + potential) of the system is conserved. As the masses oscillate, their kinetic energy is converted into potential energy as they move away from the equilibrium position, and then back into kinetic energy as they move towards the equilibrium position.

5. What happens to the oscillations if the spring is replaced with a stiffer or less stiff spring?

If the spring is replaced with a stiffer spring, the frequency of the oscillations will increase. If the spring is replaced with a less stiff spring, the frequency will decrease. The amplitudes of the oscillations may also change depending on the stiffness of the spring.

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