Period Ratio for Horizontal and Vertical Oscillations

[mbigras]

Homework Statement

A mass m rests on a frictionless horizontal table and is connected to rigid supports via two identical springs each of relaxed length l_{0} and spring constant k. Each spring is stretched to a length l considerably greater than l_{0}. Horizontal displacements of m from its equilibrium position are labeled x (along AB) and y (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 x direction.
(b) Write down the differential equiation of motion governing small oscillations in the y direction (assume y<<l).
(c) In terms of l and l_{0}, calculate the ratio of the periods of oscillation along x and y.

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

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

The Attempt at a Solution

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

(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 y so there is some force in each of the springs such that F_{L} = F_{R} = F, then the equation of motion in the y direction looks like:
<br /> 2Fsin(\theta) = ma \\<br /> sin(\theta) ≈ tan(\theta) = \frac{y}{l} \text{ *see attached attempt at inkscape*} \\<br /> 2F \frac{y}{l} = ma<br />
<br /> F = k(l&#039;-l_{0}) \\<br /> F = k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right)<br />
<br /> 2k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\<br /> 2k\left(l\left(1+\left(\frac{y}{l}\right)^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\<br /> \text{using the binomial approx...}\\<br /> 2k\left(l\left(1+\frac{1}{2}\left(\frac{y}{l}\right)^{2}\right)-l_{0}\right) \frac{y}{l} = ma\\<br /> 2k\left(y+\frac{y^{3}}{2l^{2}}-\frac{l_{0}}{l} y\right) = ma \\<br /> \text{assuming y is small} \\<br /> 2k\left(1-\frac{l_{0}}{l}\right)y = ma \\<br /> 0 = ma + 2k\left(1-\frac{l_{0}}{l}\right)y<br />
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)
<br /> \omega_{x} = \left(\frac{2k}{m}\right)^{\frac{1}{2}}\\<br /> \omega_{y} = \left(\frac{2k\left(1-\frac{l_{0}}{l}\right)}{m}\right)^{\frac{1}{2}}\\<br /> \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{-\frac{1}{2}} \\<br /> \text{The answer in the back of the book is:}\\<br /> \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{\frac{1}{2}}<br />

 

[vela]

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 $(1-l/l_0)^{+1/2}$.

 

[mbigras]

because
<br /> T_{x} = \frac{2 \pi}{\omega_{x}}<br />
right on vela, thank you.