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

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

The Attempt at a Solution

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

(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:
$$2Fsin(\theta) = ma \\ sin(\theta) ≈ tan(\theta) = \frac{y}{l} \text{ *see attached attempt at inkscape*} \\ 2F \frac{y}{l} = ma$$
$$F = k(l'-l_{0}) \\ F = k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right)$$
$$2k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\ 2k\left(l\left(1+\left(\frac{y}{l}\right)^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\ \text{using the binomial approx...}\\ 2k\left(l\left(1+\frac{1}{2}\left(\frac{y}{l}\right)^{2}\right)-l_{0}\right) \frac{y}{l} = ma\\ 2k\left(y+\frac{y^{3}}{2l^{2}}-\frac{l_{0}}{l} y\right) = ma \\ \text{assuming y is small} \\ 2k\left(1-\frac{l_{0}}{l}\right)y = ma \\ 0 = ma + 2k\left(1-\frac{l_{0}}{l}\right)y$$
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)
$$\omega_{x} = \left(\frac{2k}{m}\right)^{\frac{1}{2}}\\ \omega_{y} = \left(\frac{2k\left(1-\frac{l_{0}}{l}\right)}{m}\right)^{\frac{1}{2}}\\ \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{-\frac{1}{2}} \\ \text{The answer in the back of the book is:}\\ \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{\frac{1}{2}}$$

 

[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
$$T_{x} = \frac{2 \pi}{\omega_{x}}$$
right on vela, thank you.