# Simple Harmonic Motion Equation Question

1. Mar 16, 2008

### meganw

HELP-SEE LAST POST-I HAVE NO IDEA WHY THIS IS WRONG (SEE LAST POST)

1. The problem statement, all variables and given/known data

A spring with spring constant 230 N/m vibrates with an amplitude of 12.0 cm when 0.380 kg hangs from it.

(a) What is the equation describing this motion as a function of time? Assume the mass passes through the equilibrium point, toward positive x (upward), at t = 0.120 s.

x = A cos(omega t + phi)

A = 12 cm
W = 24.6 s^-1

Phase Shift (phi)= ___________

(b) At what times will the spring have its maximum and minimum lengths? (Consider only the first instances after t = 0.)
_____________maximum s
_____________minimum s
(c) What is the displacement at t = 0?
_____________
(d) What is the force exerted by the spring at t = 0?
____________
(e) What is the maximum speed?
_____________
When is it first reached after t = 0?
_____________

2. Relevant equations

x = A cos(omega t + phi)
$$\sqrt{k/m}$$

3. The attempt at a solution

Okay, so I've been working on this problem, and I got the first two parts pretty much no problem. A was practically given to you, and W was found using the formula $$\sqrt{k/m}$$. No problems there-I've bolded the two answers that I have correct.

Now, the next part is what really gets me. If I plug everything in to find phi, (the phase shift) for part a, the third answer, I don't know the x, and I don't know how to find it.

Thank you so much for the help!

Last edited: Mar 16, 2008
2. Mar 16, 2008

### DylanB

Assume the mass passes through the equilibrium point, toward positive x (upward), at t = 0.120 s.

you should be able to find your x and your t from this statment for starters. remember that x is the position of the mass with respect to the equilibruim position.

3. Mar 16, 2008

### meganw

Now I tried that, but it says my answer is wrong.

I set the equation equal to zero, since thats the distance from equilibrium:

note: 2.952 is 24.6*.120
0=12cos(2.952+phase shift)

My graphing calculator has a zero at 87.04 degrees, divide that by 57.3 for the radian conversion, and I get 1.52 radians, which it says is wrong. I'm sorry,-what am I doing wrong?

4. Mar 16, 2008

### DylanB

hmmm, your calculator seems to be giving an incorrect answer. the next zero should occur when the argument of the cosine function is equal to 3pi/2, and you can derive the phase shift from that. (this zero also is acceptable for the question since it is also increasing at this point).

its good to learn how to do these problems without a calculator, so you can easily check your answer, since calculators can often be unreliable due to the restrictions on the range of the inverse trig functions.

5. Mar 16, 2008

### meganw

Wait, derive the phase shift? What do you mean by "derive" the phase shift? Is my equation correct??

6. Mar 16, 2008

### DylanB

Sorry, i use the word derive too often, lol. I mean, simply calculate.

7. Mar 16, 2008

### meganw

I really don't understand this at all. I'm so sorry-but could you walk me through the steps?

8. Mar 16, 2008

### meganw

(by the way I put in 3pi/2 for the phase shift and it said that was wrong)

9. Mar 16, 2008

### Snazzy

You know x, you know A, you know omega, and you know t. You can find the phase shift using the SHM formula of that wave function.

10. Mar 16, 2008

### meganw

I did that, but I still got the wrong answer it says, assuming I have the correct (????) Equation:

0=12cos(2.952+phase shift)
phase shift=87.04 degrees, divide that by 57.3 for the radian conversion, and I get 1.52 radians

wrong , according to my webassign. :(

I let
x=0
A=12
W=24.6
T=.120

I'm sorry, I think I will be able to fly through the rest of the parts of this problem once I can get this equation finally. Thanks again. :)

Last edited: Mar 16, 2008
11. Mar 16, 2008

### DylanB

The equation is very much correct.

arccos(0) = 2.952 + phi
pi/2 or 3pi/2 = 2.952 + phi

Now can you see why the value on the left of the equation should be 3pi/2 rather than pi/2 (what's different about the cosine function at these points?)

12. Mar 16, 2008

### meganw

Why wasn't I getting that answer? Thats so weird. I had my calculator in degrees-I wonder if that was making a difference? THANK YOU!!! =)

13. Mar 16, 2008

### meganw

Wait, I tried to find the maximum and minimum and I got that answer wrong.

Isn't there a minimum at x=.056 seconds

If I'm in radians...I get a minimum at x=.056secs with this equation:

x=12cos(24.6x+1.76)

What's wrong with this??

Last edited: Mar 16, 2008
14. Mar 16, 2008

### meganw

NOTE: I tried it in degrees to, converted 1.76 to 101, and it still says I have the wrong answer=3.2 secs).

15. Mar 16, 2008

### meganw

The problem asks me to find the min/max distances from equlibrium after t=0

If I'm in radians...I get a minimum at x=.056secs with this equation:

x=12cos(24.6x+1.76)

What's wrong with this??