Simple Harmonic Motion-Vertical Spring

Hello, PhysicsIzHard.

There's something left out of this equation. Can you spot it? Also, You'll need to think carefully about the value of the phase constant θ.

Yes.

 

phi is an angle. So, it can't have units of meters.

 

The point where the mass hangs at rest before it is set in motion is the equilibrium point of the simple harmonic motion. Let this point be x = 0 and take the direction of positive x to be downward. So, at t = 0 the mass is located at x = 0 and has a positive initial velocity. You should be able to use this information to determine the phase constant.

 

The general equation for the velocity is v = -ωAsin(ωt + θ), where I've adopted your original notation of using theta for the phase angle. At t = 0, we know that the mass is passing through the equilibrium point with positive velocity. Also, the speed is a maximum when passing through the equilibrium position.

At t = 0 the velocity equation is v = -ωAsin(θ). Now, think about what the angle θ should be so that v will have it's maximum positive value.

 

180^o equal [itex]\pi[/itex] radians. But that's not the correct value. You might remember that the sine function oscillates between +1 and -1. So, v = -ωAsinθ will have its largest positive value when sinθ = -1. For what angle θ does sinθ = -1?

We are considering the instant when t=0.

 

In my previous post I pointed out that at the initial time we have v = -ωAsinθ where θ is the phase angle (which is often denoted [itex]\phi[/itex] (phi)). If you want the velocity to have its maximum negative value at the initial time, then you need to have sinθ = 1. That means that θ = [itex]\pi[/itex]/2. If you want the velocity to have its maximum positive value at t = 0, then you need to have sinθ = -1. This implies θ = -[itex]\pi[/itex]/2.

In your specific problem, you can either take upward as the positive direction or downward as the positive direction.

If you take upward as positive, then the initial velocity is negative. So, θ = [itex]\pi[/itex]/2.

If you take downward as positive, the the initial velocity is positive. So, θ = -[itex]\pi[/itex]/2.

In either case, you will get the same answers for the questions (3) and (5).

 

Well, to see that one, you should go to the equation for x: x = Acos(ωt+θ)
Letting t = 0, we have x = Acosθ. Maximum displacement in the negative direction means that x = -A. Now, x = Acosθ will reduce to x = -A when cosθ = -1. So, θ = [itex]\pi[/itex].

I'm afraid it's getting late for me, so I'm headed to bed. I will check back tomorrow. Cheers.