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A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are reflected by walls, and return. The hallway is 46.0 m long and the tuning fork is located 14.0 m from one end. What is the phase difference between the reflected waves(in degrees) when they meet at the tuning fork? The speed of sound in air is 343 m/s. [answer 68.2°] Pls show steps

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Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Step-by-step explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference


\Delta x=2(L_(2)-L_(1))

Put the value into the formula


\Delta x =2((46.0-14.0)-14.0)


\Delta x=36\ m

We need to calculate the wavelength

Using formula of wavelength


\lambda=(v)/(f)

Put the value into the formula


\lambda=(343)/(240)


\lambda=1.42\ m

We need to calculate the phase difference

Using formula of the phase difference


\phi=(2\pi)/(\lambda)* \delta x

Put the value into the formula


\phi=(2\pi)/(1.42)*36


\phi=159.29\ rad


\phi\approx 68.2^(\circ)

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

User Yahiko Kikikoto
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