I am in a terrible headache doing these problems...can someone please help me? I will post my thoughts and the calculations that I have done...

1) A 50-cm-long wire with a mass of 1.0 g and a tension of 440 N passes across the open top of a vertical tube partially filled with water. The wire, which is fixed at both ends, is bowed at the center so as to vibrate at its fundamental frequency and generate a sound wave. The water level in the tube is slowly lowered until the sound wave from the wire sets up a standing wave in the tube. It is then lowered another 36.0 cm until the next standing wave is detected. Use this information to determine the speed of sound in air.

I got that the linear density = 1x10^-3/0.5 = 2x10^-3 kg/m
And so the speed on the string is sqrt(tension/linear density) = 469.04m/s
Now how can I proceed? I looked through all examples in the textbook, but I still can't figure out what to do...


2) Two radio antennas are 100 m apart along a north-south line. They broadcast identical radio waves at a frequency of 3.0 MHz. Your job is to monitor the signal strength with a handheld receiver. To get to your first measuring point, you walk 800 m east from the midpoint between the antennas, then 600 m north. What is the phase difference between the waves at this point?
I found that the path difference is 79.964 m, and lambda = speed/frequency = 3x10^8/3x10^6 = 100m
Now what can I do? Can someone give me a hint?



Any help is appreciated! Thanks!

1) Find the fundamental frequency in the string. Because you detect standing waves only at nodes, the distance between resonances (in the tube of water) is half the wavelength.

Now you have frequency and wavelength. There is a formula that relates frequency, wavelength, and speed of propogation. Use that.

2.) I can't remember the formulae, but I think there's a simple relation between things you already have. Sorry if this isn't much help...

79.964 m / 100 m/cycle = .79964 cycle It's not usual to explicitly state 'per cycle' when expressing wavelength, but I find it usefull.

1) Can someone kindly show me how I can get the fundmental frequency and wavelength? (in which I can plug into the equation v=f(lambda) ?

I still don't get it.....help......:(

Ok. You have calculated the velocity of wave motion for the string. Now for the fundamental mode of the string the length between the ends must be 1/2 wave length. This places pressure nodes at the ends and a velocity node in the middle. So the wavelength is 2*0.5 m or 1m (100cm).

1) How can I find the fundamental frequency, then? I don't quite understand the situation pictorially...I am not sure what's happening and I don't know how to use the "36.0 cm" data...


2) How can I know the initial phase difference (specifying the initial conditions?)

Also, there is a related question: "If you now begin to walk farther north, does the signal strength increase, decrease, or stay the same?"
The correct answer is "increase", but I don't understand why...can somebody explain this?

Thanks a lot!

When a standing wave forms (like in a streched string or a pipe with one open end) you have certain boundary conditions. For a string (and for a pipe closed at _both_ ends) the ends must coinside with a node (sometimes called a pressure node) a minimum motion point. The middle is an antinode (maximum motion). The distance between two nodes is 1/2 wavelength of the wave as it travels through the string. So using the speed through the string and the length of the string you can calculate the frequency the string vibrates. Now remeber that this frequency never changes for the duration of the experiment.

Now for resonance in a one closed end pipe. The boundary condition is that a node must be at the closed end and an antinode must be at the open end. (This is where the terms pressure and velocity node come from that I used in an earlier post, the closed end will have a maximum pressure variation and the open end a maximum velocity variation.) Now we need to talk about modes. For example the string can vibrate at the fundamental n*f n=1, or at a higher mode n*f , where n = 2,3 ... For the string at n=2 from one end to the other you have node, antinode, node, antinode, node. All spaced equally. We're not doing that with the string though. Just notice how the boundary conditions are still met, nodes at the two ends of the of the sting.

Now we do something clever with the pipe. With a constant driving frequency, we are finding two different resonant legths. Our 36cm is the difference in the two lengths. The first mode for the pipe is node at the water and antinode at the open mouth. The distance between the two is 1/4 wave at the speed of sound. Unfortunately this distance isn't given. But the next mode for the single end pipe is node (at water), antinode, node, antinode (at mouth). For a total of 3/4 wavelength. 3/4 -1/4 = 1/2 wavelength and we _do_ have that difference measurement, 36cm. So now we know the wavelength of a known frequency in air.