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CioCio

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A semiclosed tube is sounded at room temperature. It is observed that the tube can be made to oscillate rather easily at several frequencies that include 1000, 1400, and 1800 Hz consecutively. What is the length of the tube?

I know that these are sequential harmonics. Therefore, given that it's a semiclosed tube and L = length of tube in meters:

λ

λ

λ

λ

_{1}= 4Lλ

_{3}= 4L/3λ

_{5}= 4L/5λ

_{7}= 4L/7I'm having trouble figuring out how to determine exactly

*which*harmonics these are. Converting the frequencies into their respective periods gives me:

T

T

T

T

_{n}= [itex]\frac{1}{1000 Hz}[/itex] = 0.001000 s

T

_{n+2}= [itex]\frac{1}{1400 Hz}[/itex] = 0.0007143 s

T

_{n+4}= [itex]\frac{1}{1800 Hz}[/itex] = 0.0005555 s

(where n is the n

^{th}harmonic)

Looking at the basis for harmonics (in semiclosed tubes), I know that

*if a wave travels a distance of 4L, or an odd-numbered integral quotient of 4L, in a time equal to its own period, a standing wave is formed.*In other words, the 1000 Hz wave listed above must travel 4L/n meters in 0.001000 second for a standing wave to form.

But how can I determine what n is? One cannot simply assume that 1000 Hz is the fundamental frequency. Likewise, as shown by the basic wave formula of v = ƒλ, one cannot calculate wavelength from frequency/period alone.

I thought about an approach based around velocity but, again, I only know the time component of one oscillation.

Assuming room temperature (23 °C; 296 K), I know that:

**ƒ**

_{n}= [itex]\frac{86n}{L}[/itex]The calculation would be simple, but I don't know how I can calculate n. I apologize for not having more work to show, and I've tried to make up for that by posting my logic, but I'm at a standstill.