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

• lizzyb
In summary: So if you want to find the phase difference between two waves with a known phase difference, just divide the known phase difference by 360 and use the result as the argument in the sine or cosine function.
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??

??

lizzyb said:
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

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!

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

conejoperez28 said:
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°"

## 1. How does a tuning fork produce sound?

A tuning fork produces sound when it is struck against a surface or when it vibrates due to an external force. The two prongs of the fork vibrate at a certain frequency, creating sound waves that travel through the air and are perceived as sound by our ears.

## 2. What is the purpose of a tuning fork in music?

Tuning forks are used to produce a specific, consistent pitch that can be used as a reference for tuning musical instruments. They are also used by singers to help them find the correct pitch before singing.

## 3. How is the frequency of a tuning fork determined?

The frequency of a tuning fork is determined by the length and thickness of its prongs. The longer and thicker the prongs, the lower the frequency, and vice versa. The frequency is also affected by the material and temperature of the tuning fork.

## 4. Can tuning forks be used for anything other than music?

Yes, tuning forks have various applications in science and medicine. They are commonly used in hearing tests, as well as in resonance experiments to study the properties of sound waves. They are also used in watchmaking to regulate the timekeeping of mechanical watches.

## 5. How do you tune a tuning fork?

Tuning forks can be tuned by filing the ends of the prongs to adjust their length and thickness. They can also be tuned by heating or cooling them to change their frequency. Advanced techniques such as laser trimming can also be used to precisely tune a tuning fork to a specific frequency.

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