A) \[666 Hz\]
B) \[753 Hz\]
C) \[500 Hz\]
D) \[333 Hz\]
Correct Answer: A
Solution :
[a]Frequency of the sound produced by open flute \[f=2\left( \frac{v}{2\ell } \right)=\frac{2\times 330}{2\times 0.5}=660Hz\] |
Velocity of observer, \[{{\operatorname{V}}_{0}}=10\times \frac{5}{18}=\frac{25}{9}m/s\] |
As the source is moving towards the observer therefore, according to Doppler?s effect. |
\[\therefore \]frequency delected by observer, \[f'=\left[ \frac{v+{{v}_{0}}}{v} \right]f=\left[ \frac{\frac{25}{9}+330}{330} \right]660\] |
\[=\frac{2995}{9\times 330}\times 660\operatorname{or},f'=665.55=666Hz\] |
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