Mass Connected to two springs

In summary, the problem involves a mass connected to two springs on a frictionless table. The differential equations of motion for small oscillations in the x and y directions are derived, and the ratio of the periods of oscillation along these directions is calculated in terms of the relaxed length and spring constant. The correct answer is found to be proportional to (1-l/l_0)^{+1/2}.
  • #1
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Homework Statement


A mass [itex]m[/itex] rests on a frictionless horizontal table and is connected to rigid supports via two identical springs each of relaxed length [itex]l_{0}[/itex] and spring constant [itex]k[/itex]. Each spring is stretched to a length [itex]l[/itex] considerably greater than [itex]l_{0}[/itex]. Horizontal displacements of [itex]m[/itex] from its equilibrium position are labeled [itex]x[/itex] (along AB) and [itex]y[/itex] (perpendicular to AB).

*see attached image or link*

(a) Write down the differential equiation of motion (i.e., Newton's law) governing small oscillations in the [itex]x[/itex] direction.
(b) Write down the differential equiation of motion governing small oscillations in the [itex]y[/itex] direction (assume [itex]y[/itex]<<[itex]l[/itex]).
(c) In terms of [itex]l[/itex] and [itex]l_{0}[/itex], calculate the ratio of the periods of oscillation along [itex]x[/itex] and [itex]y[/itex].

The rest of the question can be found on page 86 at http://www.scribd.com/doc/160672855/Vibrations-and-Waves-a-P-French



Homework Equations


[tex]
F = ma \\
a^{2} + b^{2} = c^{2} \\
sin(\theta)≈tan(\theta) \text{ when } \theta \text{ is small} \\
(1+a)^{n} ≈ 1 + na \frac{n(n-1)a^{2}}{2!} +...\\
[/tex]



The Attempt at a Solution


(a) I imagined a small displacement [itex]x[/itex] to the right, which I also declared to be positive.
[tex]
-k(l + x - l_{0}) + k(l-x-l_{0}) = ma \\
-2kx = ma \\
0 = ma + 2kx
[/tex]

(b) I tried going about this using some of the techniques discussed while doing question 3-7 which can be found here: https://www.physicsforums.com/showthread.php?t=694259 But my answer doesn't match the one in the back of the book and I'm not sure why.

I also imagined a small displacement [itex]y[/itex] so there is some force in each of the springs such that [itex]F_{L} = F_{R} = F[/itex], then the equation of motion in the [itex]y[/itex] direction looks like:
[tex]
2Fsin(\theta) = ma \\
sin(\theta) ≈ tan(\theta) = \frac{y}{l} \text{ *see attached attempt at inkscape*} \\
2F \frac{y}{l} = ma
[/tex]
[tex]
F = k(l'-l_{0}) \\
F = k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right)
[/tex]
[tex]
2k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\
2k\left(l\left(1+\left(\frac{y}{l}\right)^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\
\text{using the binomial approx...}\\
2k\left(l\left(1+\frac{1}{2}\left(\frac{y}{l}\right)^{2}\right)-l_{0}\right) \frac{y}{l} = ma\\
2k\left(y+\frac{y^{3}}{2l^{2}}-\frac{l_{0}}{l} y\right) = ma \\
\text{assuming y is small} \\
2k\left(1-\frac{l_{0}}{l}\right)y = ma \\
0 = ma + 2k\left(1-\frac{l_{0}}{l}\right)y
[/tex]
Also I see in the last step there is something going on with a sign, maybe it's because the way I set it up the displacement is in the negative direction.

(c)
[tex]
\omega_{x} = \left(\frac{2k}{m}\right)^{\frac{1}{2}}\\
\omega_{y} = \left(\frac{2k\left(1-\frac{l_{0}}{l}\right)}{m}\right)^{\frac{1}{2}}\\
\frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{-\frac{1}{2}} \\
\text{The answer in the back of the book is:}\\
\frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{\frac{1}{2}}
[/tex]
 
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  • #2
You just made an algebra mistake in the last step. The ratio ##\frac{T_x}{T_y} = \frac{\omega_y}{\omega_x}## has ##\omega_y## on top so it should be proportional to ##(1-l/l_0)^{+1/2}##.
 
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  • #3
because
[tex]
T_{x} = \frac{2 \pi}{\omega_{x}}
[/tex]
right on vela, thank you.
 

1. How does the mass affect the frequency of oscillation?

The mass does not affect the frequency of oscillation in a system connected to two springs. The frequency is determined by the stiffness of the springs and the mass has no effect on this value.

2. What is the relationship between the mass and the amplitude of oscillation?

The mass and the amplitude of oscillation have an inverse relationship. As the mass increases, the amplitude of oscillation decreases. This is because the heavier mass requires more force to displace it from its equilibrium position.

3. Can the mass be varied without changing the frequency of oscillation?

Yes, the mass can be varied without changing the frequency of oscillation in a system connected to two springs. The frequency is dependent on the stiffness of the springs, not the mass.

4. How does the position of the mass affect the tension in the springs?

The position of the mass does not affect the tension in the springs. The tension is determined by the stiffness of the springs and remains constant regardless of the position of the mass.

5. What happens to the frequency of oscillation if one of the springs is removed?

If one of the springs is removed, the frequency of oscillation will decrease. This is because the stiffness of the system has decreased, leading to a slower oscillation. The frequency can also change depending on how the remaining spring is connected to the mass.

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