# Homework Help: Simple Harmonic Oscillation on a guitar string

1. Jan 26, 2012

### kgal

1. The problem statement, all variables and given/known data
A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 mm and a phase angle of zero.
a. Write an equation for the position of the center of the string as a function of time.
b. What are maximum values of the magnitude of the velocity and the acceleration of the center of the string?
c. The derivative of the acceleration with respect to time is a quantity called jerk. Write an equation for the jerk of the center of the string as a function of time and find the maximum value of the magnitude of the jerk.

2. Relevant equations
x(t) = Acos(ωt + ρ)
ω = 2∏f = v/r

3. The attempt at a solution
a. I found that x(t) = Acos(ωt + ρ) = Acos(ωt).
b. d/dt x(t)= dx/dt = v(t) = -Aωsint(ωt) = 0
dv/dt = a(t) = Aω^2cos(ωt) = 0
I know that you need to set the equations equal to zero in order to find maximum magnitudes but I'm stuck from there on.

c. da/dt = jerk = -Aω^3sin(ωt) = 0.
same problem for me as in section b where I had trouble understanding how I find the max. jerk.

2. Jan 26, 2012

### Staff: Mentor

The sine and cosine functions, all by themselves, both have maximum amplitudes. What are they? What does that say about the maximum or minimum values that the velocity and acceleration functions can have?

3. Jan 26, 2012

### kgal

maximum of sine and cosine functions are 1. so do I set up cos/sin(wt) = 1?

4. Jan 26, 2012

### TheLil'Turkey

No, because that's incorrect. But x max = A, v max = Aω, etc.

5. Jan 26, 2012

### Staff: Mentor

Only the sine and cosine functions are varying in your expressions for position, velocity and acceleration. They are multiplied by constants. If you know the maximum values that the sine and cosine functions can have, then you should be able to say what the minimums and maximums will be for the position, velocity and acceleration without doing any calculus.

6. Jan 26, 2012

### kgal

got it, thanks!

7. Dec 17, 2012

### leonardthecow

I realize this is an old thread, but I'm working on this same problem now. I understand the equations and have the solutions to each part, but what I don't understand is the signs in the answers. For instance, in part A), the correct answer is given as x(t)=.003cos(-2764.6t). It's easy enough to write in the amplitude and find omega, but why is omega negative here? Further, when determining the equation for the jerk, the correct answer is given as j(t)=(6.3x10^7)sin(2764.6t). How is it that omega now became positive through the process of taking derivatives?? I'm trying to work on these wave functions for my final exam, help would be much appreciated

8. Dec 17, 2012

### Staff: Mentor

There's no need to plug in values to find ω, since ω can be found directly from the given frequency of vibration, 440 Hz. $\omega = 2\pi f$.

Keep in mind that for functions of the form A*sin(t), that A is the magnitude of an oscillation, so the function will take on values between the extremes -A to +A. When you take the derivative and the resulting function ends up with a negative sign in front of it (because, for example, the derivative of Bcos(ωt) is -ωBsin(ωt)), then the amplitude is ωB; Amplitudes are positive values. The "-" sign tells you the relationship between the phase of the derivative sin() function and the original Bcos() function. Note that $-ω B sin(ω t) = ω B sin(ω t + \pi)$, so the "-" sign can be easily interpreted as a 180° phase shift.

9. Dec 17, 2012

### leonardthecow

I should have specified, this example comes from the Mastering Physics website, and is an old problem set I am reworking for practice. The correct answer that they identify is x(t)=.003cos(-2764.6t), or x(t)=.003cos[-440(2pi)t].

What I was originally confused about was the fact that they identify the correct omega to have a minus sign in front of it. I used the general equation x(t)=Acos(ωt+φ), with a phase shift in this case of 0. Plugging in f, ω=2π(440 Hz); there is no negative sign here, nor was it clear to me where it came from (when I originally entered x(t)=.003cos(880π), the website marked it as incorrect, with the message "check your signs".) This seems to imply that the answer takes the form of x(t)=Acos(-ωt), and I'm unsure of why that is. Am I misunderstanding your response?

Thank you for your explanation of signs in relation to phases, that hadn't explicitly occurred to me before and it helps to clarify things conceptually. However, I'm still not sure in this case why the sign in front of ω is affected, not the sign in front of A. Thank you

10. Dec 17, 2012

### Staff: Mentor

Well then, it's a mystery to me, too. I can't think of any good physical reason to introduce what are essentially either negative frequencies or negative times. Both seem distinctly unphysical things for introductory classical mechanics, and as far as I can tell are unwarranted for this problem.

Of course, mathematically cos(-θ) = cos(θ) so it makes no difference...

11. Dec 17, 2012

### leonardthecow

Thanks for all your help!! Knowing I'm not totally losing it then helps as much as a physical answer would. I guess by virtue of the definition that cos(-θ)=-cos(θ), maybe they're assuming that the guitar string is initially being released from the negative direction of maximum displacement? Not sure why... Regardless, thank you for your time