# Homework Help: Pipe and speed of sound question

1. Aug 23, 2011

### xtheunknown0

1. The problem statement, all variables and given/known data
A student holds a 325.6 Hz tuning fork near the end of a pipe that is closed at one end. He notices that resonances can be heard for several different lengths, one of which is 55.5 cm. He gradually lengthens the pipe to 92.5 cm where he detects the next resonance.

2. Relevant equations

3. The attempt at a solution
Let lamda_1 = 0.555 * 4 = 2.22 m
f_1 = 325.6 Hz
v = 723 m/s

lamda_3 = 4/3 * 0.925
v = 402 m/s

Why aren't the two answers consistent?

2. Aug 23, 2011

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

How do you know that 325.6 Hz is the fundamental mode?

3. Aug 23, 2011

### cmb

Because you've not figured out the right numbers, maybe...?

What is the Highest Common Factor of 55.5 and 92.5?

4. Aug 23, 2011

### Hootenanny

Staff Emeritus
Or the GCD of 555 and 925, as 55.5 and 92.5 don't have a GCD.

5. Aug 23, 2011

### PeterO

The reason we do the experiment like this is due to end correction. [look it up]

A pipe that is 55.5 cm long "behaves" as if it a little longer. The 92.5 cm pipe also behaves as if it also a little longer. Fortunately the extra length each time [the end correction] is the same [it is related to the cross-sectional area of the pipe - or in the case of a round pipe, a constant*radius.

The shortest stopped pipe that exhibits resonance behaves as if it is 1/4 wavelength long [it will physically be a little shorter]. The next length to exhibit resonance is behaving as 3/4 wavelength, then 1,25 , 1.75, 2.25 etc.

So the additional length each time is exactly half a wavelength.

That should see you right.

NOTE: had this been an open pipe, the pipe would exhibit resonance when behaving as if it was 0.5, 1, 1.5, 2, 2.5, 3, etc wavelengths long. Same method of solution used.

That is why this sort of experiment is done by lengthening the pipe rather than increasing the frequency to get other resonances.

EDIT: There is always a chance that this "experiment" has been idealised and end correction was assumed to be zero.

6. Aug 23, 2011

### cmb

OK, fair enough. I should've gone for suggesting a Lowest Common Multiple directly (2.775m) and worked back, or somehow worded it as 'a common lowest-integer multiplicand' (18.5).

7. Aug 23, 2011

### PeterO

As I explained earlier - the key to this experiment is that the additional length to get to the next resonance is half a wavelength. End of story.

8. Aug 24, 2011

### cmb

As I didn't directly explain earlier (folks can do their homework), get the factor of two resonances right and you've got the 1/4 wavelength (then, don't even need to know if they are 'immediate' resonances, or the 'next' one). End of Appendix.