Question about this simple harmonic motion problem

[jklops686]

Homework Statement

A particle with a mass of 65 g is moving with simple harmonic motion. At time t = 0, the particle is at its extreme positive displacement of 18.0 cm. The period of the motion is 0.600 s. Find the vecocity of the particle at t = 1.35 s

Homework Equations

(1). ω=2∏/T or
(2). ω=√k/m

(3). x=Acos(ω∏+δ)
(4). v=-Aωsin(ω∏+δ)

The Attempt at a Solution

First question: to find ω, why don't I get the same answer for both equations 1 and 2?

With equation 1 I get 10.47 (which I think is correct) and with eqn. 2 I get 7.45.

Second question: How do I find the phase change for this problem? I tried to set the displacement equation equal to zero at t=0 and I am getting ∏/2 but it seems that the only way to get the correct answer for the problem is to have zero phase change.

 

[TSny]

First question: to find ω, why don't I get the same answer for both equations 1 and 2?

With equation 1 I get 10.47 (which I think is correct) and with eqn. 2 I get 7.45.

It will help if show your calculations. In particular, how did you determine the value of k to use in eqn. 2?

Second question: How do I find the phase change for this problem? I tried to set the displacement equation equal to zero at t=0 and I am getting ∏/2 but it seems that the only way to get the correct answer for the problem is to have zero phase change.

The problem states that the particle is at its extreme positive displacement at t = 0.

 

[SteamKing]

How do you know what k is?

 

[jklops686]

How do you know what k is?

I was thinking I could use hooke's law to find k. Tsny...

I guess I'm just confused at finding phase changes.

 

[TSny]

What does the equation x = Acos(ωt+δ) become for t = 0? Knowing that x is at it's maximum positive value at t = 0, what can you conclude about the value of δ?

 

[SteamKing]

But how did you get a value of k to calculate omega of 7.45?

 

[jklops686]

But how did you get a value of k to calculate omega of 7.45?

Using F=kx

0.65N=k(.18)

k=3.61

ω=√3.61/.065=7.45



TSny... I think I see what you're saying. x would equal 18 at t=0.

so 18cm = Acos(δ) ?


Also, how do you quote two people in one reply?

 

[TSny]

Using F=kx

0.65N=k(.18)

k=3.61

ω=√3.61/.065=7.45

But you can't assume that the force acting on the particle is the weight of the particle. The simple harmonic motion might be due to a spring, and the force would be whatever the force of the spring happens to be for a given value of x.

x would equal 18 at t=0.

so 18cm = Acos(δ) ?

What does "A" stand for in this equation? What is the value of A?

Also, how do you quote two people in one reply?

Pause your cursor over the "M" button next to the "Quote" button in the lower right corner and instructions will appear.