A 1 lb weight is suspended from a spring. Let y give the deflection (in inches) of the weight from its static deflection position, where “up” is the positive direction for y. If the static deflection is 24 in, find a differential equation for y. Solve, and determine the period and frequency of the SHM of the weight if it is set in motion.
Finding the period and frequency for simple harmonic motion
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ordinary-differential-equations
discrete-mathematics
physics
1 Answers
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Well, $F=kx$ for spring. Since static equilbrium balances against gravity $kx=mg$. On the other hand, when in motion the constant force of gravity does not effect the oscillation frequency and we can write $m\ddot{y} +ky=0$ which gives characteristic equation $m\lambda^2+k=0$ hence $\lambda = \pm i \sqrt{k/m}$. Let $\omega = \sqrt{k/m}$ and find $y(t) = A\cos(\omega t + \phi)$ as the solution to the equations of motion. The period $T$ is defined such that $\omega T = 2\pi$ hence $T = 2\pi /\omega$. You are given $k(24)=1$ in inches and lbs. To finish you need to deal with the difference between $m$ and $mg$. Beware the inches vs. ft issue. Good luck. (if you believe in that sort of thing)
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0I'm curious as to how the condition $k(24) = 1$ allows me to to come up with a solution to the differential equation. – 2012-10-25