JEE Advanced 2010Paper 1
- Exam
- JEE Advanced
- Year
- 2010
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Waves and Sound
JEE Advanced 2010 · Paper 1 · Physics · Waves and Sound
A stationary source is emitting sound at a fixed frequency f_0 ,...
A stationary source is emitting sound at a fixed frequency , which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is of . 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 .
Correct answer
0
Step-by-step 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, 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 $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$ $\therefore$ The nearest integer is 7 . Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$\begin{aligned} f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\ \therefore \quad 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] \\ \text{ 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 \end{aligned} As and are very very less than . We can write, \left(v-v_1\right) or \begin{aligned} & =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\ & =7.128 \mathrm{kmh}^{-1} \end{aligned} The nearest integer is 7 .
Hence, 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 $\left(v_1-v_2\right)=\frac{v \times 1.2}{200}$ $\therefore$ The nearest integer is 7 . Hence, $\quad f_1=f_0\left(\frac{v+v_1}{v-v_1}\right)$\begin{aligned} f_2 & =f_0\left(\frac{v+v_2}{v-v_2}\right) \\ \therefore \quad 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] \\ \text{ 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 \end{aligned} As and are very very less than . We can write, \left(v-v_1\right) or \begin{aligned} & =\frac{330 \times 1.2}{200}=1.98 \mathrm{~ms}^{-1} \\ & =7.128 \mathrm{kmh}^{-1} \end{aligned} The nearest integer is 7 .