What Is the Phase Difference at the Tuning Fork in a Hallway Sound Experiment?

[lizzyb]

A tuning fork generates sound waves with a frequency of 246 Hz. The waves travel in opposite directions along a hallway, are reflected by end walls, and return. The hallway is 47.0 m long, and the tuning fork is located 14.0 m from one end. What is the phase difference between the reflected waves when they meet at the tuning fork? The speed of sound in air is 343 m/s.

We have the equation: \Delta r = \frac{\phi}{2 \pi} \lambda so it seems that all we need to do is determine phi since we can easily determine delta r and lambda. But the answer that I come up with is different than in the book.

\lambda = \frac{v}{f} = \frac{343}{246} = 1.39 m

Easy enough. But what about the change in r? Let r1 be the distance traveled by the sound that goes to the left and r2 be the sound that goes to the right, thus we have:

r_1 = 2(47 - 14) = 66 m
r_2 = 2(14) = 28 m
\Delta r = r_1 - r2 = 38 = \frac{\phi}{2 \pi} \lambda
so \frac{38 \cdot 2 \cdot \pi}{1.39} = \phi

Which is 171.77 radians maybe? But this is way off the answer in the back of the book, 91.3 degrees, because 171.77 * 180 / pi = 9841.7 degrees modulo 360 = 121 degrees??

??

 

[Andrew Mason]

A tuning fork generates sound waves with a frequency of 246 Hz. The waves travel in opposite directions along a hallway, are reflected by end walls, and return. The hallway is 47.0 m long, and the tuning fork is located 14.0 m from one end. What is the phase difference between the reflected waves when they meet at the tuning fork? The speed of sound in air is 343 m/s.

We have the equation: \Delta r = \frac{\phi}{2 \pi} \lambda so it seems that all we need to do is determine phi since we can easily determine delta r and lambda. But the answer that I come up with is different than in the book.

\lambda = \frac{v}{f} = \frac{343}{246} = 1.39 m

Easy enough. But what about the change in r? Let r1 be the distance traveled by the sound that goes to the left and r2 be the sound that goes to the right, thus we have:

r_1 = 2(47 - 14) = 66 m
r_2 = 2(14) = 28 m
\Delta r = r_1 - r2 = 38 = \frac{\phi}{2 \pi} \lambda
so \frac{38 \cdot 2 \cdot \pi}{1.39} = \phi

Which is 171.77 radians maybe? But this is way off the answer in the back of the book, 91.3 degrees, because 171.77 * 180 / pi = 9841.7 degrees modulo 360 = 121 degrees??

??

The phase difference is the path difference minus the number of whole wavelengths in the path difference.

The number of wavelengths in 38.0 m is 38.0/(343/246) = 27.254. The number of whole wavelengths is 27 so the phase difference is .254 of a wavelength or 360 x .254 = 91.3 degrees. To the correct significant figures, the phase difference is really .3 of a wavelength or 108 degrees.

AM

 

[lizzyb]

wow - my error was in rounding off the results. thanks so much. As a recap, I can come up with the correct answer with:

\phi = \frac{\Delta r \cdot 2 \cdot \pi}{\lambda} = \frac{\Delta r \cdot 2 \cdot \pi}{v/f} = \frac{\Delta r \cdot 2 \cdot \pi \cdot f}{v} = \frac{(66 - 28) \cdot 2 \pi \cdot 246 }{343} radians \cdot \frac{180 degrees}{\pi radians}

anway, i came up with 9811.31195335 modulo 360 = 91.3 degrees!

 

[conejoperez28]

I don't understand the "modulo" conversion, its not working for me, can you explain it please?

 

[OlderDan]

I don't understand the "modulo" conversion, its not working for me, can you explain it please?

The sine and cosine functions are periodic with a period of 360°. A "phase difference" of n*360 + θ is the same as a phase difference of θ for all integer values of n. When you have some big angle and subtract n*360 from it, you are finding the angle "modulo 360°"