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  4. A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2 m / s one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be: (given that velocity of sound =300 m / s ; g=10 m / s 2 )

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Q. A sounding body emitting a frequency of $150\, Hz$ is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of $2\, m / s$ one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be: (given that velocity of sound $=300\, m / s ; g=10\, m / s ^{2}$ )

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Solution:

$f=f_{0}\left(\frac{v \pm v_{0}}{v \pm v_{s}}\right)$
When approaching:
$f_{a}=150\left[\frac{300+2}{300-10}\right]$
When receding:
$f_{r}=150\left[\frac{300-2}{300+10}\right]$
$\Rightarrow f_{a}=f_{r}=12$