Wave Motion 5 Question 23
23. A stationary source is emitting sound at a fixed frequency $f _0$, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is $1.2 %$ of $f _0$. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is $330 ms^{-1}$.
(2010)
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Answer:
Correct Answer: 23. 7
Solution:
- Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.
Hence, $f _1=f _0\left(\frac{v+v _1}{v-v _1}\right)$
$$ f _2=f _0\left(\frac{v+v _2}{v-v _2}\right) $$
$\therefore f _1-f _2=\left(\frac{1.2}{100}\right) f _0=f _0\left[\frac{v+v _1}{v-v _1}-\frac{v+v _2}{v-v _2}\right]$
or $\left(\frac{1.2}{100}\right) f _0=\frac{2 v\left(v _1-v _2\right)}{\left(v-v _1\right)\left(v-v _2\right)} f _0$
As $v _1$ and $v _2$ are very very less than $v$.
We can write, $\left(v-v _1\right)$ or $\left(v-v _2\right) \approx v$
$\therefore \quad\left(\frac{1.2}{100}\right) f _0=\frac{2\left(v _1-v _2\right)}{v} f _0$
or $\quad\left(v _1-v _2\right)=\frac{v \times 1.2}{200}=\frac{330 \times 1.2}{200}=1.98 ms^{-1}$
$=7.128 kmh^{-1}$
$\therefore$ The nearest integer is 7 .
