#$&*
course phy 202
Find the frequencies of the first four harmonics of a standing sound wave in an open pipe of length 3 meters and in which sound propagates at 340 m/s.¥This corresponds to 1/4 wavelength, so the wavelength of this harmonic must be 4 * 4 meters = 16 meters.
¥ The second harmonic goes from node to antinode to node to antinode, a distance of 3/4 wavelength. Thus 3/4 wavelength = 4 meters and the wavelength is 4/3( 4 m ) = 5.3 meters.
¥ Similarly the third and forth harmonics span 5/4 and 7/4 of a wavelength, respectively, and there respective wavelengths are thus 4/5( 4 m) = 3.2 m and (4/7)( 4 m) = 2.2 m.
Since sound moves 340 m in each second, there will be 340/ 16 = 21.25 peaks in each second of the fundamental harmonic. The frequency is therefore 21.25 Hz.
Similarly the three overtones will have frequencies of
¥ ( 340 m/s) / ( 5.3 m) = 64.2 Hz,
¥ ( 340 m/s) / ( 3.2 m) = 106.3 Hz
¥ ( 340 m/s) / ( 2.2 m) = 154.5 Hz
**** I believe I did this correct, I used an intro problem for guidance, does this look right?
@&
This is an open pipe.
Its harmonics have node-antinode structure
A N A
A N A N A
A N A N A N A
etc..
Respectively, these structures correspond to 1/2, 1, 3/2, ... wavelengths.
So the 3 meter length of the pipe will be 1/2, 1, 3/2, ... wavelengths. It follows that the wavelengths will be 2, 1, 2/3, ... times the length of the pipe.
The ratios you are using correspond to those for a pipe which is closed on one end, open on the other. This would result in node-antinode structures
N A
N A N A
N A N A N A
etc., in which the length of the pipe corresponds to 1/4, 3/4, 5/4, ... wavelengths. The result is that the wavelengths are respectively 4, 4/3, 4/5, ... times the length of the pipe.
The main point is that if you try to solve the problem without starting with the node-antinode structure, you are much less likely to end up with a valid solution.
*@
What is the angle of total internal reflection from a material which has index of refraction 1.6 to a material with index of refraction 1.03?
arcsin(1.03/1.6)= 40.07 deg
arcsin((sin(40.07 deg)*1.6/1.03)= 86.86 deg
angle of total internal reflection= about 90 degrees
Problem Number 1
A string of length 8 meters is fixed at both ends. It oscillates in its third harmonic with a frequency of 199 Hz and amplitude .31 cm. If it is held under a tension of 9 Newtons, then what is its mass?
**** I am having trouble with this one for some reason, could you help me with an equation? do I use this equation? v = sqrt( T / (mass/unitLength) )
@&
The formula relates T, mass per unit length and propagation velocity v.
The problem gives you T.
To find the mass you also need the propagation velocity.
You get this by the fact that the string oscillates in the third harmonic at 199 Hz.
What is the node-antinode structure of the third harmonic?
How far is it between a node and an antinode?
What therefore is the wavelength?
What given quantity do you combine with the wavelength to find the propagation velocity?
&&&&& I still do not understand this problem
I know propagation velocity formula, I just do not know how to find the wavelength?
@&
The string is fixed at both ends. So there are nodes at both ends.
The possible node-antinode structures for a string with nodes at both ends are
N A N
N A N A N
N A N A N A N
etc.. (the above being the first three harmonics).
The reasoning is then the same as in my preceding note.
See what you can do with this and get back to me.
*@
Problem Number 4
A sound source with frequency 130 Hz approaches an observer at 95 m/s. If the speed of sound is 340 m/s, then what frequency will be heard by the observer?
fl=((v+vl)/v)*fs=((340+95)/340)*130
=166.3 Hz
"