What is the expression for vertical velocity in a simple harmonic oscillator?

[nissanztt90]

Homework Statement

Vertical equation for motion:

$$\frac{d^2z}{dt^2} +v^2 = 0$$

which corresponds to a simple harmonic oscillator, with an angular frequency $$v = \sqrt{4 \pi G \rho}$$. The solution of the diff eq can be written as $$z(t) = Asin(vt+\phi_0)$$

where z(t) is the vertical position of a test particle, A is the amplitude of its motion, and $$\phi_0$$ is its initial phase.

Given z(t), write out the expression for the vertical velocity.

Homework Equations

The Attempt at a Solution

Since z(t) is the vertical position, the derivative of z(t) with respect to t would be the velocity? Correct?

And the vertical velocity expression would be $$\frac{dz}{dt} = Acos(vt+\phi_0) * v$$? Is this correct?

 

[Redbelly98]

Yes, that's correct.

Just as an aside, most people and most textbooks use ω for angular velocity, so instead of "vt" those terms would be ωt.

Also, I think the original equation is

$$\frac{d^2z}{dt^2} +v^2z = 0$$

 

[nissanztt90]

Yes your correct about the original equation...also not sure why they used v for the angular velocity.