# Wave Changing Air Pressure (should be easy, help please!)

1. Nov 27, 2005

### Seiya

. [Tipler5 15.P.049.] A sound wave in air produces a pressure variation given by the following formula, where p is in pascals, x is in meters, and t is in seconds.

p(x,t) = 0.70 cos /2 (x - 330t)

(a) Find the pressure amplitude of the sound wave.
correct check mark Pa
(b) Find the wavelength.
wrong check mark m
(c) Find the frequency.
wrong check mark Hz
(d) Find the speed.
m/s

The only part i got is (a) which is obvious 0.70....

i cant get the other ones and i only have 1 more go at them. I tried frequency = w=2(pi)f so w/2(pi) = f... but that didnt work... also 330 by itself didnt work? i tought the frequency in a wave was determined by the number before t? =\ Im really confused and i read over the book 3 times already and i cant figure it out? (theres only half a page on pressure and waves)

Any help appreciated... thanks a lot!

2. Nov 27, 2005

### lightgrav

you should edit your post so others know that it is "cos pi/2 ...".

As in oscillations, the frequency is the reciprocal of the repeat time.
cosine repeats each 2 pi ... if t = 0 at x=0, cos = 1
if t = 1/330 [sec], at x=0 you have cos (0+ pi/2) = 0 .
if t = 2/330 [sec], at x=0 you have cos (0+ pi) = -1 .
If t = 4/330 [sec], at x=0 you have cos (0+ 2 pi) = 1 .
. . . the repeat time is 4/330 .
We do this "properly" by remembering the solution "cos (omega t)"
which means that omega is (pi/2)330 .

In a similar way, the wave number "k" is in solution "cos (kx)"
Don't forget to distribute the pi/2 (that is, multiply thru by it).

Speed is a different matter. There we only need the argument of the function,
writing it so it looks like (x - vt)/lamda ... or as ... k (x - vt) .

enough?

3. Nov 28, 2005

### Seiya

thanks a bunch. =)