A string of length 6 meters is fixed at both ends. It oscillates in its third harmonic with a frequency of 147 Hz and amplitude .81 cm. What is the equation of motion of the point on the string which lies at 1.6 meters from the left end? What is the maximum velocity of this point?
I remember doing this in the intro problems like this but I seem to be having trouble with it.
In a nutshell you find the amplitude of the oscillation at that point using the function amplitude(x) = A sin( 2 pi / lambda * x).
The simple harmonic motion of the point is then y = amplitude(x) sin( omega * t), where omega is the angular frequency of the oscillation.
If this doesn't refresh you memory, go back to the Intro Problem Sets where a complete and detailed solution to this type of problem is presented. You may of course ask questions about that solution.
Find the frequencies of the first four harmonics of a standing wave in a uniform string of length 5 meters which is fixed at both ends and in which a wave propagates at 42 m/s
1st Harmonic-
Length of string will be twice the length (5m) = 10m
freq. = v/lambda
42 m/s / 10 m
= 4.2 Hz
2nd-
Length of string will be L, or the same length, 5m
42 m/s / 5 m
= 8.4 Hz
3rd
Length of String will be 2/3 L, or 5m * 2/3 = 3.3 m
42 m/s / 3.3 m
= 12.7 Hz
4th
Confused on what the length would be for the 4th harmonic. What would the length be, what would I multiply the length 5m by (what fraction), would it be 4/3 ?
The condition on this string is that it has nodes at both ends. So the node-antinode configuration can be N A N, or N A N A N, or N A N A N A N, etc., corresponding to 2, 4, 6, ... node-antinode distances.
Since the distance between a node and an antinode is 1/4 wavelength, this means that the length of the string can be 2 * 1/4 wavelength, 4 * 1/4 wavelength, 6 * 1/4 wavelength, ... , i.e., 1/2, 2/2, 3/2, ... wavelengths. So what would be the next couple of numbers in the sequence 1/2, 2/2, 3/2, ... ?
So the wavelengths can be 2/1, 2/2, 2/3, ... times the length of the string. You have figured out that since the length of the string is 5 m, the first three wavelengths would be 2/1 * 5 m = 10 m, 2/2 * 5 m = 5 m, 2/3 * 5 m = 3.3 m. What would be the next few ratios after 2/1, 2/2, 2/3 ? What therefore would be the next couple of wavelengths?
What the first four fundamental frequencies of a transverse standing wave in an aluminum rod of length 3 meters, balanced at its midpoint, if a transverse disturbance travels in the rod at 1500 m/s? Where should the rod be supported in order to permit the second harmonic to ring freely?
Would I go about doing this problem just as the one before? How would I find where the rod should be supported to ring clearly, I am confused on this part of the question.
The aluminum rod would have antinodes at its ends, because the ends are free. The node-antinode configurations of the first few harmonics are A N A, A N A N A, A N A N A N A, etc. The analysis of wavelengths and frequencies uses the same reasoning as before.
For the A N A configuration the node is at the middle, and if you support a rod at any point but a node the vibrations will be eliminated by the support (at a node the oscillation of the rod has zero amplitude--i.e., it isn't moving at that point--so it's no problem to prevent motion at that point by putting the support there).
In the second harmonic the configuration is A N A N A. Where on the rods would the nodes be? That's where it has to be supported.
If a traveling wave has wavelength 2.7 meters and the period of a cycle of the wave is .071 second, what is the propagation velocity of the wave? What is its frequency?
Period = lambda/ v
v = lambda/period
= 2.7 m / .071 sec
= 38 m/s
Freq. = v / lambda
38 m/s / 2.7 m
= 14 Hz
Good.
A sound source with frequency 360 Hz moves away from an observer at 85 m/s. If the speed of sound is 340 m/s, then what frequency will be heard by the obsever?
Confused on where to start this problem…How far away did the source originate, how can I find that out?
I would need this to find the pulses emitted which would lead me to the frequency.
You can choose any distance you like, and you can use any time interval you like. For a source moving away from the observer, it's probably easiest to start with the source right at the observer. It's easiest to think in terms of a time interval over which a significant number of cycles occur. A very convenient time would be 1 second, which would work find because here 360 cycles will occur in 1 second.
How many of these 360 cycles will have reached the observer at the end of a second, and how many will still be in transit between the source and the observer?
Traveling waves are set up in a pair of long strings by a single harmonic oscillator. The strings have identical mass densities of 6 grams / meter. Both strings terminate at a short bungee cord attached to a wall. The harmonic oscillator is attached to the other ends of the strings in such a way that one string is 7.2 meters longer than the other. If the oscillator has frequency 63 Hz:
• Give at least two string tensions which will produce the maximum motion of the bungee cord.
• Give at least two string tensions which will produce the minimum motion of the bungee cord.
I am blank on where to go with this problem, I reviewed the intro problems/notes but they are not helping me very much.
You really need to explain specifically what you do and do not understand about the corresponding intro problem. Otherwise all I can do is risk confusing the issue with a different type of explanation, or repeat the given explanation.
Here's my best attempt:
Maximum discomfort will occur if antinodes from one string meet antinodes from the other, which will happen if the 7.2 meters corresponds to 1, 2, 3, ... wavelengths. If the 7.2 meters corresponds to 1 wavelength, what is the propagation velocity? What therefore must be the tension (remember that propagation velocity = sqrt( T / (m / L) ) ).
Answer the same for 2 wavelengths and you'll have two tensions.
Minimum discomfort occurs if antinodes meet nodes, which occurs if the 7.2 meters is 1/2, 3/2, 5/2, ... wavelengths (i.e., a whole number plus a half wavelength). The reasoning is otherwise the same as the above.
Sound is created by the vibration of the air column in a pipe which is open at both ends. The pipe is 5.9 meters long, and the speed of sound is 348 m/s. What are the frequencies of the first four natural harmonics? What are the successive frequency ratios (i.e., the ratio of each but the first frequency to the next lower frequency), and how well can each be approximated by a power of 2^(1/12)?
1st natural harmonic- 2L = 5.9 *2 = 11.8 m
348 m/s / 11.8 m = 29.5 Hz
2nd – 3/2 L = 5.9 m * 3/2 = 8.9 m
348 m/s / 8.9 m = 39.1 Hz
3rd - 4/3 L = 5.9 * 4/3 = 7.9
348 m/s / 7.9 m = 44.1 Hz
4th – 5/4 L = 5.9 * 5/4 = 7.4 m
348 m/s / 7.4 m = 47 Hz
I am not sure what you mean by successive freq. ratios, do you mean like how the 3/2 and 4/3 ratios are approximated very closely by the 7th and 5th powers of 2^(1/12)?
That is exactly what the question means. From your question it appears that you pretty much have the answer; let me know if not.