1. Tardigrade
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  4. A source and an observer move away from each other, each with a velocity of 10 m/s, with respect to ground. If the observer finds the frequency of sound coming from the source as 1950 Hz, the original frequency of the source is: (assume, velocity of sound in air = 340 m/s)

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Q. A source and an observer move away from each other, each with a velocity of 10 m/s, with respect to ground. If the observer finds the frequency of sound coming from the source as 1950 Hz, the original frequency of the source is : (assume, velocity of sound in air = 340 m/s)

EAMCETEAMCET 1997

Solution:

Applying Dopplers effect, if source and observer both are moving away from each other. Then, the apparent frequency heard by observer is given by $ n=\left( \frac{v-{{v}_{o}}}{v+{{v}_{s}}} \right)n $ $ \left( \begin{align} & Given:n=1950\,Hz \\ & {{v}_{o}}={{v}_{s}}=10\,m/s \\ & v=340\,m/s \\ \end{align} \right) $ $ 1950=\left( \frac{340-10}{340+10} \right)n $ $ \Rightarrow $ $ n=\frac{1950}{\frac{340-10}{340+10}}=\frac{1950\times 350}{330} $ $ =2068\,Hz $