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.
To my understanding, the initial condition is y(0) = 24. I have a differential equation of the form $my'' + ky = 0$, and I know that m = 1, but how do I find k?
Best Answer
Hint: The characteristic equation for your differential equation is: $$ m\lambda^2 + k = 0$$ which is equivalent to $ \lambda^2 + \frac{k}{m} = 0$. Therefore, $\lambda = \pm \sqrt{\frac{k}{m}}$. And hence the solutions for the differential equation are: $$ y(x) = \alpha_1\cos \left (\sqrt{\frac{k}{m}}x \right) + \alpha_2\sin \left (\sqrt{\frac{k}{m}}x \right) $$ where $\alpha_i \in \mathbb{R}$.