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A 500-Hz whistle is moved toward a listener at a speed of 10.0 m/s. At the same time, the listener moves at a speed of 20.0 m/s in a direction away from the whistle. What is the apparent frequency heard by the listener? (The speed of sound is 340 m/s.)

Respuesta :

Answer:

f'  = 485 Hz

Explanation:

given,

Frequency of whistle,f = 500 Hz

speed of source, v_s = 10 m/s

Speed of observer, v_o - 20 m/s

speed of sound,v = 340 m/s

Apparent frequency heard = ?

Using Doppler's effect formula to find apparent frequency

[tex]f' = (\dfrac{v-v_0}{v-v_s})f[/tex]

[tex]f' = (\dfrac{340-20}{340-10})\times 500[/tex]

[tex]f' = 0.9696\times 500[/tex]

f'  = 485 Hz

Hence, the apparent frequency is equal to 485 Hz.

Lanuel

To determine the apparent frequency heard by the listener is equal to 545.45 Hz.

Given the following data:

  • Observer velocity = 20.0 m/s
  • Frequency of sound = 500 Hz
  • Source velocity = 10.0 m/s
  • Speed of sound = 340 m/s

To determine the apparent frequency heard by the listener, we would apply Doppler's effect of sound waves:

Mathematically, Doppler's effect of sound waves is given by the formula:

[tex]F_o = \frac{V \;+ \;V_o}{V\; - \;V_s} F[/tex]

Where:

  • V is the speed of a sound wave.
  • F is the actual frequency of sound.
  • [tex]V_o[/tex] is the observer velocity.
  • [tex]V_s[/tex] is the source velocity.
  • [tex]F_o[/tex] is the apparent frequency.

Substituting the given parameters into the formula, we have;

[tex]F_o = \frac{340 \;+ \;20}{340\; - \;10} \times 500\\\\F_o =\frac{360}{330} \times 500\\\\F_o =1.0909 \times 500[/tex]

Apparent frequency = 545.45 Hz.

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