# A 342-g mass is attach to a spring and undergoes SHM.

1. Nov 29, 2016

### utorontoconnor

1. The problem statement, all variables and given/known data
A 342-g mass is attach to a spring and undergoes SHM.
Its max accelerations is 18.6m/s^2
Its max velocity is 1.75m/s

2. Relevant equations
x(t)=Acos(wt)
x'(t)=-wAsin(wt)
x''(t)=-w^2Acos(wt)
w=sqrt(k/m)
3. The attempt at a solution
I essentially derived these equations and have no idea where to go or how to equate them... Using Wolfson "University Physics" 3rd edition.

2. Nov 29, 2016

### Staff: Mentor

What is the question?

3. Nov 29, 2016

### utorontoconnor

Oops sorry... sequentially determine A) the angular frequency B) the amplitude, and C) the spring constant

4. Nov 29, 2016

### eys_physics

For simple harmonic motion the general solution is $x(t)=Acos(\omega t+\phi)$ where $\omega=\sqrt{k/m}$ and $\phi$ is a phase. I don't know why you are assuming that $\phi=0$. Please, also give the problem statement. What is the question?

5. Nov 29, 2016

### utorontoconnor

Terribly sorry... I'm not sure what my first steps would be to find the angular frequency, ω, given a mass of 342-g attached to a spring undergoing simple harmonic motion, a max acceleration of 18.6m/s^2, and a maximum speed* of 1.75 m/s. Essentially, I don't know how to use F=-kx, x(t)=Acos(ωt + φ) and the derivatives to solve the aforementioned, angular frequency.

6. Nov 29, 2016

### Staff: Mentor

Since you're given the maximum values for the acceleration and velocity and you weren't given a particular starting position, you don't need to worry about a phase angle. You can assume that the phase angle is zero. Your set of relevant equations is fine.

From your relevant equations, pick out the expressions that represent the maximum acceleration and velocity.

7. Nov 29, 2016

### utorontoconnor

x'(t)=-wAsin(wt)
So, if I want to maximize my velocity, I'd need to make sin(wt)=1, since sine oscillates between -1, and 1, the appropriate time would be at π/2? Hence, I would be left with x'(t)max=-wA(1)
then, x''(t)=-w^2Acos(wt) would need to have cos(wt)=1 as well in order to get max velocity, because it's acceleration is greatest when x is further away form equilibrium?
If so, then x''(t)=-w^2A

which then leads me to where...? Sorry I'm have a tough go at wrapping my head around this. I'm sure its simpler than I'm making it out to be.

8. Nov 29, 2016

### Staff: Mentor

Yes, the coefficient of the sine or cosine sets the maximum amplitude of the term since the maximum value for sine or cosine is 1.

You've been given values for the maximum acceleration and velocity. Given them variable names so you don't have to drag around a lot of digits. Maybe call them Amax and Vmax. Write out the expressions for Amax and Vmax that you've extracted from your relevant equations.

9. Nov 29, 2016

### utorontoconnor

Vmax= -ωA
Amax=-ω^2A
can I then equate ω=Vman/A? and the sub that into my Amax ω?

10. Nov 29, 2016

### utorontoconnor

Or am i able to isolate for A and then replug that into either equation to solve for ω?

11. Nov 29, 2016

### Staff: Mentor

Yes.
Either way, it's up to you. The important thing is that you have a path forward towards your goal.

Note that you can drop the negative signs from the terms since they represent the maximum values, hence are absolute values (and you could have chosen sine and cosine to be -1 rather than +1 to make those expressions positive).

You might want to try out the x2 and x2 icons in the edit window for producing subscripts

12. Nov 29, 2016

### utorontoconnor

Thank you! Really appreciate the guidance... Although, I appear to be off by a factor 1/10:
Vmax=wA
Amax=w^2A
where w= Vmax/A, thus Amax=[(Vmax/A)^2]xA
Hence, A= Vmax^2/Amax= 3.0625/18.6= .165m.
Hence, w=1.75/.165= 10.60... though my text says it should be 1.06 or 10.6^-1...
Where would my error be for that?

13. Nov 29, 2016

### Staff: Mentor

I agree with your result. Confirm that the given values are correct.

Textbooks have been known to have typos and errors, particularly if problems are "updated" for new editions and the answer key is not updated, too. For example, if Amax happened to be 1.86 instead of 18.6 m/s2 then that could account for the difference.

See how the value you get for the spring constant compares with what they say it should be.

14. Nov 29, 2016

### utorontoconnor

Sorry- I think the textbook just has some typos...

15. Nov 29, 2016

### utorontoconnor

Sorry, that was a waste of your time! Thanks again for all of your help! I look forward to your help again, hopefully!
Cheers

16. Nov 29, 2016

### Staff: Mentor

No worries. We're here to help