Simple Harmonic Oscillator and period

[Knissp]

[SOLVED] Simple Harmonic Oscillator

Homework Statement

The equation of motion of a simple harmonic oscillator is (second derivative of x wrt t) d2x/dt2 = -9x, where x is displacement and t is time. The period of oscillation is?

Homework Equations

2 pi f = omega
f = 1/T

The Attempt at a Solution

Given the relevant equations, one can show that
omega / 2pi = f
2pi/omega = T

The answer is 2pi/3 by the way, and I have trouble getting to this number. I probably have to integrate something, but I don't know what that would buy me. dx/dt = -9/2 x^2 and x = -3/2 x^3. When x is sinusoidal, I know the general form of the equation, but I am totally lost here.

 

[mikelepore]

The sine will go away when you cancel.

x = A sin omega t
v = dx/dt = A omega cos omega t
a = dv/dt = d2x/dt2 = -A omega^2 sin omega t

The problem given: d2x/dt2 = -9x
becomes: -A omega^2 sin omega t = -9 (A sin omega t)

everything cancels except: - omega^2 = -9
omega = 3

omega = 2 pi f = 2 pi / T
3 = 2 pi / T

T = 2 pi / 3

 

[Moetasim]

The period of a simple harmonic oscillator can be determined using the equation T = 2π/ω, where ω is the angular frequency. To find ω, we can use the given equation of motion and substitute in the values for x and t. This will result in a differential equation that can be solved for ω. The solution to this differential equation is ω = 3, which means that the period of oscillation is T = 2π/3. This is consistent with the given answer of 2π/3. Therefore, the period of the simple harmonic oscillator is 2π/3.