Whooping C. Crane

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Resonant frequency physics question? The windpipe of a typical whooping crane is about 5.3 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 35°C.) The speed of sound in dry air [m/s] is approximately 331.4 + (0.6 * temperature in degreesC). It is slightly higher for air with water vapor as is the case of the throat of the whooping crane, but not much higher.
A pipe with a closed end and an open end will resonate at wavelength 4 times the length of the pipe.
5.3 feet * 4 = 21.1 feet = 6.43 meters
speed of sound in air at 35C = 331.4 + (0.6 * 35) = 352.4 m/s
352.4 m/s divided by 6.43 m = 54.8 Hz
(Hertz is the same unit as "per second")
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Finding Frequency...how do you do this problem? The windpipe of a typical whooping crane is about 5.3 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 34°C.) windpipe with one-end closed
always >>> open end forms A (antinode)
closed end >> forms N (node)
O open ----- L ------ * closed
A ................ ........... N
for lowest frequency = entire length will have one loop (now intermediate)
L = distance between A and N = λ / 4
λ = 4L
L = 5.3 feet = 5.3*12*2.54/100 = 1.61544 meter
freq (funda or lowest) = n = v/ λ = (331.3 + 0.6*t) / 4L
n = (331.3 + 0.6*t) / 4*1.61544
n = 54.43 Hz

How do you do this question? The windpipe of a typical whooping crane is about 4.8 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 35°C. Given L=4.8 ft = 1.46m
At 35'C, speed of sound is: v=352 m/s
lowest resonant freq: n=1
The resonant frequency of a closed cylinder (pipe) is:
f = nv/4L = 60 Hz

An ecologist counting the whooping cranes in a wildlife refuge over a five-year period is studying ecology at? a. organism. b. population. c. community. d. ecosystem. b, population

can you help me with this simple programming questions? (arrays)? Lesson 13 1. Write code that will declare a double array called sgt having 800 elements. 2. double []dfw = new double[21]; System.out.println( dfw.length ); //What is printed? 3. For the code in #1 above, write a for-loop that will cycle through all the elements of double sgt[] and store the square root of the index of each element in that element. 4. Assuming rtl is the name of an array, what’s wrong with this code? double rtl_len = rtl.length( ); 5. On one line of code, both declare a character array called cr and initialize its elements to be ‘a’, ‘b’, ‘c’, ‘d’, and ‘e’. 6. Refer to #5 above. What is the value of cr.length ? 7. Write code that will print the sum of the squares of the all elements of the ref integer array. 8. What’s wrong with the following code? for (int k = 2; k < homer.length; k++) { homer[k +1] = k; } 9. Fill in the blanks below to enable us to pass a double array called dbx to a method called heroWorship. Within the method, the array should be called vb. boolean bbc = heroWorship( ); public boolean heroWorship( ) //signature of method 10. Assume the five values an integer array adc contains are: 34, 56, -102, 18, and 5. What is the value of adc[1] ? 11. Using the adc array from #10 above, what would be the value of adc[3] + adc[4]? 12. Using the adc array from #10 above, what would be the value of adc[5] ? 13. Describe what the following code segment does: for (int j = 0; j < b.length; j++) b[j] = Math.abs( b[j] ); 14. For the int array c = {1, 2, 3, 4}, what would be the output of the following code? String ss = “>>>”; int len = ss.length( ); for (int j = 0; j < len; j++) ss+= c[j]; System.out.println(ss); 15. String wc = “Whooping crane”; String sp[] = wc.split(“oo”); for(int j = 0; j < sp.length; j++) { System.out.println(sp[j]); } 16. List the elements of String [ ]sArray = “fee fi fo ”.split(“\\s+”);. 17. List the elements of String [ ]sp = “One two”.split(“Q”);. 18. Using the split method, write code that will count all of the occurrences of “th” (without regard to upper or lower case) in “The best THERE is is Barth”. I am in my fourth and final year of my Computer Systems and Software Engineering course. Programming has got very little to do with your grasp of the subject, but it does make you a better developer and ready for industry. You should get out your keyboard and have a go.
Try it yourself then ask for help on the ones you have no clue about.

Can someone help me with a physics problem? The windpipe of a typical whooping crane is about 4.9 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 35°C.) The lowest frequency corresponds to a quarter of the wavelength being the length of the windpipe. If that assumption is correct (it has been about two decades since I did this :-)), the wavelength is 4 times 4.9 feet.
Figure out the speed of sound from the temperature - the formula is a standard one. Then divide by the wavelength and you have the answer.
If you have difficulty visualizing this, imagine that you are holding a tuning fork over a mouth of a vertical pipe filled with water, and the water being slowly let out through the bottom - at one point, you will hear the (first) resonant mode.

Quick Physics question? The windpipe of a typical whooping crane is about 4.8 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 35°C. λ = 4 . L = 19.2 ft = 5.85 m
Speed of sound at 35˚C = 352.5
Fundamental frequency = 362.5 / 5.85 = 61.9 Hz = 62 Hz

Physics - Sound (lowest resonant frequency)? The windpipe of a typical whooping crane is about 5.5 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 33°C.) L = lambda/4
lambda =4(5.5) = v / (f 1) = 351.63 / (f 1)
f 1 = 351.63 / 22
f 1= 15.98 Hz
speed of sound at this temp = 351.63 m/s

Please! Desperately need help!? The windpipe of a typical whooping crane is about 4.8 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 35°C. λ = 4 . L = 19.2 ft = 5.85 m
v = 352.5 m/s
f = v / λ = 60.2 Hz

Can someone help with the physics question? The windpipe of a typical whooping crane is about 4.9 feet long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? (Assume a temperature of 35°C.) Would have to know how big around the windpipe is roughly. The diameter of the pipe has just as much to do with the frequency as the length does, think about pipes of a pipe organ.
The material of the pipe may affect the tone as well, flesh probably has different resonant properties than say steel or wood.
From there, you should be able to find calculators and formulas online to get your answer.
Hope this helps! :-)

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