# Homework Help: Maximum vibration speed of a traveling wave

1. Nov 30, 2014

1. The problem statement, all variables and given/known data
y(x,t) = (9.00cm)sin((5Πcm-1)x + (3Π rad/s)t - Π/8)
There are 11 parts, and I have answered the first 10 (they include velocity of a wave, amplitude, all that good stuff), but I haven't seen anything in lecture or in the textbook that mention finding maximum vibration speed.

2. Relevant equations
y(x,t) = ymsin(kx - wt + Φ)

2. Nov 30, 2014

### haruspex

What do you think it means by 'vibration speed'? It could mean frequency, but then it would not be asking for a maximum. What else could it be?

3. Nov 30, 2014

The only thing I could think of was taking the derivative of the original equation (not sure whether with respect to x or t) and then setting equal to zero and solving for v (probably with respect to x then) but I would still be stuck with another variable.

4. Nov 30, 2014

### haruspex

You didn't answer my question. There's no point guessing at procedures until you know what it is you're trying to find.

5. Nov 30, 2014

Ok then, I see what you mean. I believe it means the maximum speed of the element of string. So the speed as the element is at y = 0.

6. Nov 30, 2014

### haruspex

Yes.

7. Nov 30, 2014

Ok so, to find this speed, I would take the derivative of the function as one would do when doing this for oscillations, correct? I just am not sure how to do this with variables x and t inside the cosine. Or am I going the wrong direction?

8. Nov 30, 2014

### haruspex

y is a function of two independent variable, x and t: y = y(x,t). What direction does an element move in? Algebraically, what expression represents the speed of an element at position x and time t?

9. Nov 30, 2014

dx/dt would, if that's what you are asking...but it moves in the y direction. I'm a bit confused on that...

10. Nov 30, 2014

### haruspex

The elements are not moving in the x direction. The wave moves in the x direction, but we're not concerned here with the speed of the wave.
Exactly.

11. Dec 1, 2014

So the velocity would be dy/dt, and to get this, take the derivative with respect to t, getting:

dy/dt = 27π(cos((5πcm-1)x + (3π rad/s)t - π/8)

This gives velocity (but still has x and t in it). Wouldn't I have to set the second derivative equal to zero to find the maximum velocity?

12. Dec 1, 2014

### haruspex

You could do that, but as you already said it will occur when y = 0. What does that tell you about the argument of the cosine function in your dy/dt formula?

13. Dec 1, 2014

Well if you were to use the second derivative, the sin function would equal 0, so maybe the cosine function equals 1?

14. Dec 1, 2014

### RUber

If the max is when cosine equals 1, what is that velocity? It looks like you are already done; just make sure your units are correct.

15. Dec 1, 2014

### haruspex

Yes, but as I wrote, you can get that from y = 0, without taking derivatives.

16. Dec 1, 2014

I understand what you are asking, but I don't understand how to obtain any information from this. From dy/dx = 27π(cos((5πcm-1)x + (3π rad/s)t - π/8) I can't seem to draw any numerical information. Could I get a hint to get in the right direction?

17. Dec 1, 2014

### haruspex

You mean dy/dt, right?
At y(x,t) = 0 you have (9.00cm)sin((5Πcm-1)x + (3Π rad/s)t - Π/8) = 0. What does that tell you about sin((5Πcm-1)x + (3Π rad/s)t - Π/8) at that time and position? And what does that tell you about cos of the same argument? So what do you deduce for dy/dt?

18. Dec 1, 2014

That would tell me that sin-1(0) = 5πcm-1x + (3π rad/s)t - π/8 .
This in turn would tell me that π/8 = 5πcm-1x + (3π rad/s)t and that 1/8 = 5cm-1x + (3rad/s)t. I'm just not sure what direction I should be going, what I am truly solving for.

19. Dec 1, 2014

### haruspex

No, it's simpler that that. If sin(theta) = 0, what can you say about cos(theta)?

20. Dec 1, 2014

cos(theta) would equal 1 then, so the maximum vibration speed would be equal to 27π cm/s or rad/s (not quite sure on the units anymore lol).

21. Dec 1, 2014

### haruspex

Yes.
y is in cms, t is in secs, so dy/dt is cm/s.

22. Dec 1, 2014

Awesome. So replacing that junk with 0, which I "kind of" solved for/set equal to, allows for me to use 0 in the dy/dt equation which gives dy/dt = 27π(cos(0)), which is equal to 27π. Seems logical.

23. Dec 2, 2014

### RUber

Think about what the units are telling you. You really have 9cm $\times 3\pi$ rads per second. If $3\pi$ rads per second means 1.5 full cycles per second, and a cycle is from +9cm to -9cm, what is the speed of the string?
I could be wrong, but that is the only way this makes sense to me.

24. Dec 2, 2014

### haruspex

Not sure what you are suggesting as the answer. Radians are dimensionless. If an arc of a circle radius 1cm subtends an angle of 2 radians at the centre, the length of the arc is 2cm, not 2cm-radians.