Question

a simple harmonic oscillator has amplitude A and time period T what is the maximum speed of it?

Answer 1

Answer:

A simple harmonic oscillator is one which has a restoring force which acts so as to restore the system back towards the equilibrium point.

[math]F=ma=m\frac{d^2x}{dt^2}=m\ddot{x}=-kx[/math]

The solutions to this form of equation are sinusoidal.

[math]x(t)=Acos(\omega t+\phi)[/math]

This is a standard result. You can easily see this by “trying” a solution, however you can mathematically find this by using known methods of solving second-order linear differential equations.

To find the velocity we need to differentiate [math]x(t)[/math]

as [math]v(t)=\frac{dx(t)}{dt}[/math]

so [math]v(t)=-A \omega sin(\omega t + \phi)[/math]

Given we want to find the maximum velocity, and [math]A[/math] and [math]\omega[/math] are constant, maximum velocity or speed will be when [math]sin(\omega t+\phi)[/math] is maximum, which is when [math]sin(\omega t+ \phi) = 1[/math]

Therefore, maximum velocity [math]v_{max}=|-A\omega|=A\omega[/math]

If we follow the same method, we can find the maximum acceleration to be

[math]a_{max}=A\omega^2[/math]

as [math]a([/math][math]t)=\frac{dv(t)}{dt}[/math]

so [math]a(t)=A [/math][math]\omega^2 cos(\omega t + \phi)[/math]

and [math]cos(\omega t + \phi) = 1[/math] when at maximum.