Mass Connected to two springs

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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]
 

Answers and Replies

  • #2
vela
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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
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because
[tex]
T_{x} = \frac{2 \pi}{\omega_{x}}
[/tex]
right on vela, thank you.
 

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