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Tanya [424]
3 years ago
15

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
Physics
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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Therefore, the original length of the simple pendulum is 2.97 m

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