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Introductory Physics Homework Help
Period Ratio for Horizontal and Vertical Oscillations
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[QUOTE="mbigras, post: 4490204, member: 485229"] [h2]Homework Statement [/h2] 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]. [I]Each spring is stretched to a length [itex]l[/itex] considerably greater than [itex]l_{0}[/itex][/I]. 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 [i]periods[/i] 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 [h2]Homework Equations[/h2] [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] [h2]The Attempt at a Solution[/h2] (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: [URL="https://www.physicsforums.com/showthread.php?t=694259"]https://www.physicsforums.com/showthread.php?t=694259[/URL] 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] [/QUOTE]
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Introductory Physics Homework Help
Period Ratio for Horizontal and Vertical Oscillations
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