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scZoUnD [109]
2 years ago
9

A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound s

ource. A student walks at a uniform rate of 1.12 m/s away from one speaker and toward the other. How many beats does the student hear per second? (Take the speed of sound to be 343 m/s.) Hint: The Doppler effect causes both frequencies to be shifted. The difference between those two frequencies is what causes the beats.
Physics
1 answer:
vampirchik [111]2 years ago
6 0

Answer:

2.97795 Hz

Explanation:

v = Speed of sound in air = 343 m/s

v_r = Relative speed between the speakers and the student = 1.12 m/s

f' = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz

Frequency of sound heard as the student moves closer to the other speaker

f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz

The difference in the frequencies is

f_2-f_1=457.48897-454.51102=2.97795\ Hz

The student hears 2.97795 Hz

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The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>
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  • Final velocity (v) = 0 m/s (at maximum height)
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The maximum height to which the ball can attain can be obtained as follow:

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Divide both side by –19.6

h = –64 / –19.6h

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Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

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initial speed, u = 0 ft/s

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