Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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)

share|cite|improve this answer
So the differential equation I need to solve is simply $my'' + ky = 0; y(24) = 1$? – user1038665 Oct 25 '12 at 3:24
I'm curious as to how the condition $k(24) = 1$ allows me to to come up with a solution to the differential equation. – user1038665 Oct 25 '12 at 6:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.