# Homework Help: Wave equation confusion!

1. Mar 25, 2006

### mr_coffee

Hello everyone! I'm having troubles understanding why i'm not getting this right...the problem is:

A sinusoidal wave is traveling on a string with speed 10. cm/s. The displacement of the particles of the string at x = 25 cm is found to vary with time according to the equation y = (5.0 cm) sin[10.0 - (4.0 s-1)t]. The linear density of the string is 3.0 g/cm.

What is the frequency of the wave?
Well wouldn't just be 4.0 s^-1?

Also it says What is the wavelength of the wave?
I t hought it would be:
You know the wave number is: (5.0 cm) and you know the relationship:
Wave Length = 2PI/k, where k is the wave number: so i put:
Wave Length = 2PI/5 = 1.2566 which was wrong any ideas why i'm not getting these right?

THanks!

2. Mar 25, 2006

### Galileo

Find the period T first. After one period, the phase has advanced by $2\pi$. Then use f=1/T.

The wavenumber is not 5.0 cm, that's the amplitude.
Use $\lambda f=v$ to find the speed of the wave.

3. Mar 25, 2006

### Staff: Mentor

4.0 is the angular frequency of the wave, $\omega$, in radians/sec; not the frequency, $f$, in cycles/sec. What do you need to multiply radians/sec by, in order to get cycles/sec?

4. Mar 25, 2006

### mr_coffee

Thanks for the responce! I'm still kind of confusd on how you find the period with the given information, i'm very rusty on trig. I understand sin has a period of 2PI, but that doesn't mean that function's period is just 1/2Pi does it?

I alwyas found hte period by first finding the angular frequency w, which is suppose to be the number infront of the t. So in this case, would it be 1/4.0, since it says 4.0s^-1?
or is it suppose to be w = -1/4.0? Then i can find the period by using:
f = w/2*pi
T = 1/f

oo my bad, the wave number should be what is infront of the x, but there is no x in the equation :\
I really don't have anything to look at as a reference other then the internet becuase my book doesn't have chapter 16, yes!

Didn't see your message until now jtbell, well 2PI = 360 degree's, so i'm asumming, i would take 4.0 rad/sec *360/2PI ?
But it does say s^-1, so again does this mean its angular frequency is 1/4?

Last edited: Mar 25, 2006
5. Mar 25, 2006

### Galileo

Right. The angular frequency $\omega$ is 4.0/s. And this is related to the frequency by $f=2\pi \omega$.

You can get this simply by considering when the phase advances by $2\pi$. Consider $\sin(\omega t+\phi)$. One period passes if $\omega T=2\pi$ or $T=2\pi/\omega=1/f$.

And don't fuss with degrees. Use radians.

6. Mar 25, 2006

### mr_coffee

Thanks for the help!
f = w/2*pi i think though!
but i got it right once i flip it around. Now i found all the parts to the question but i'm confused on how i'm suppose to find the tension of this string.
I found the following:
What is the frequency of the wave?
.63662
What is the wavelength of the wave?
15.708
Give the general equation giving the transverse displacement of the particles of the string as a function of position and time.
y(x,t) = (5cm)*sin[ (.3999cm^-1)x -(4s^-1)t]
which was correct. Now they said:
What is the tension in the string?

I don't have the book chapter, and in the notes it shows no equation dealing with tension at all. Any ideas? THanks!

7. Mar 25, 2006

### Galileo

Yes, you're right

A disturbance in a stretched string satisfies the wave equation (for small amplitudes). The speed is:

$$v=\sqrt{\frac{T}{\mu}}$$.
Where T is the tension in the string and $\mu$ the mass per unit length.

Last edited: Mar 25, 2006
8. Mar 25, 2006

### mr_coffee

Thanks again!
hm..i don't see where i'm messing this one up at:
v = 10cm/sec
T = ?
$$\mu= 3.0 g/cm$$
if they want newtons i converted mu to kg/m
3.0 g/cm (1 kg/1000g) (100cm)/1 m = .3 kg/m;
v = .1 m/sec
v^2/$$\mu$$ = T;
T = .1^2/.3 = .03333N

but that was wrong, i also tried 333.33N by forgetting to convert the 10cm/sec to .1m/sec, any ideas where i f'ed it up?

Oops i'm a tard, it should be v^2*mu

Yep that fixed it! w00t!

Last edited: Mar 25, 2006