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3 years ago

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A source emits sound with a frequency of 860 Hz. It is moving at 20.0 m/s toward a stationary reflecting wall. If the speed of s

ound is 343 m/s, what frequency does an observer riding with the source hear

2 answers:

olga_2 [115]3 years ago

4 0

Answer:

Explanation:

f = 860 Hz

velocity of source, vs = 20 m/s

velocity of sound, v = 343 m/s

use the formula of doppler's effect

the frequency reflected from the wall is

$f' = \frac{v}{v+ v_{s}}\times f$

$f' = \frac{343}{343+20}\times 860$

f' = 812.62 Hz

Frequency heard by the observer

$f'' = \frac{343}{343-20}\times 812.62$

f'' = 862.93 Hz

Galina-37 [17]3 years ago

3 0

Answer:

$f_o=860\ Hz$

Explanation:

Given:

original frequency of sound wave, $f=860\ Hz$

speed of the sound source, $v_s=20\ m.s^{-1}$

original speed of sound wave from the source, $s=343\ m.s^{-1}$

<u>According to the Doppler's effect:</u>

$\frac{f_s}{f_o} =\frac{s+v_s}{s-v_o}$

The sound is reflected from the wall and the source is moving towards the wall and observer is also riding the same source.

The velocity of the observer, $v_o=-20\ m.s^{-1}$

$\frac{860}{f_o} =\frac{s+20}{s-(-20)}$

$f_o=860\ Hz$

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