Respuesta :
Answer:
(A) 0.773 m
(B) f' = 452.94 Hz
(C) [tex]\lambda' = 0.751\ m[/tex]
(D) f" = 427.058 Hz
(E) [tex]\lambda' = 0.796\ m[/tex]
Solution:
As per the question:
Frequency of the sound produced, f = 440 Hz
Speed of the sound in still air, v = 340 m/s
Now,
(A) To calculate the wavelength of the sound wave:
We use the relation:
[tex]v = f\lambda [/tex]
[tex]\lambda = \frac{340}{440} = 0.773\ m[/tex]
(B) By using Doppler effect to calculate the frequency of the sound wave:
Velocity of the receiver, [tex]v_{R} = 10.0\ m/s[/tex]
Velocity of the source, [tex]v_{S} = 0\ m/s[/tex]
When the receiver is approaching:
[tex]f' = \frac{v + v_{R}}{v}f = \frac{340 + 10}{340}\times 440[/tex]
f' = 452.94 Hz
(C) To calculate the wavelength of the sound wave:
[tex]\lambda' = \frac{v}{f'} = \frac{340}{452.94} = 0.751\ m[/tex]
(D) While moving away, the frequency of the sound wave can be calculated as:
[tex]f" = \frac{v - v_{R}}{v}f = \frac{340 - 10}{340}\times 440[/tex]
f" = 427.058 Hz
(E) The wavelength can be given by:
[tex]\lambda" = \frac{v}{f"} = \frac{340}{427.058} = 0.796\ m[/tex]
