Two engines are crossing each other which are travelling in opposite directions at $72kmh{{r}^{-1}}$. One engine sounds a whistle of frequency $1088cps$. What will be the frequencies as audible by an observer on the other engine before and after crossing? Assume that the speed of sound is $340m{{s}^{-1}}$.
Answer
641.4k+ views
Hint: The apparent frequency can be found by taking the product of the actual frequency and the ratio of the resultant of velocity of sound and velocity of observer to the resultant of the velocity of sound and velocity of the source. Find the apparent frequency before and after the crossing. This will help you in answering this question.
Complete step by step solution:
The apparent frequency of the engines before crossing can be found by taking the product of the actual frequency and the ratio of the sum of velocity of sound and velocity of observer to the difference of the velocity of sound and velocity of the source. That is we can write that,
${n}'=n\left( \dfrac{v+{{v}_{0}}}{v-{{v}_{s}}} \right)$
Here in the question it has been mentioned that,
The velocity of the observer and source respectively are,
$\begin{align}
& {{v}_{0}}=72kmh{{r}^{-1}}=20m{{s}^{-1}} \\
& {{v}_{s}}=72kmh{{r}^{-1}}=20m{{s}^{-1}} \\
\end{align}$
The actual frequency will be given as,
$n=1088$
The speed of the sound will be,
$v=340m{{s}^{-1}}$
Substituting the values in it,
${n}'=1088\left( \dfrac{340+20}{340-20} \right)=1088\times \dfrac{360}{320}=1224Hz$
Now the apparent frequency after crossing will be found by taking the product of the actual frequency and the ratio of the difference of velocity of sound and velocity of observer to the sum of the velocity of sound and velocity of the source. That is we can write that,
${n}''=n\dfrac{v-{{v}_{0}}}{v+{{v}_{s}}}$
Substituting the values in it will give,
${n}''=1088\times \dfrac{340-20}{340+20}=1088\times \dfrac{32}{36}=967.11Hz$
Hence the question has been answered.
Note:
The Doppler Effect is defined as phenomena in which the variation in frequency of a wave in relation to an observer who is in motion with respect to the wave source. This phenomenon is named after the Austrian physicist Christian Doppler. He is the first one who completely explained this phenomenon in 1842.
Complete step by step solution:
The apparent frequency of the engines before crossing can be found by taking the product of the actual frequency and the ratio of the sum of velocity of sound and velocity of observer to the difference of the velocity of sound and velocity of the source. That is we can write that,
${n}'=n\left( \dfrac{v+{{v}_{0}}}{v-{{v}_{s}}} \right)$
Here in the question it has been mentioned that,
The velocity of the observer and source respectively are,
$\begin{align}
& {{v}_{0}}=72kmh{{r}^{-1}}=20m{{s}^{-1}} \\
& {{v}_{s}}=72kmh{{r}^{-1}}=20m{{s}^{-1}} \\
\end{align}$
The actual frequency will be given as,
$n=1088$
The speed of the sound will be,
$v=340m{{s}^{-1}}$
Substituting the values in it,
${n}'=1088\left( \dfrac{340+20}{340-20} \right)=1088\times \dfrac{360}{320}=1224Hz$
Now the apparent frequency after crossing will be found by taking the product of the actual frequency and the ratio of the difference of velocity of sound and velocity of observer to the sum of the velocity of sound and velocity of the source. That is we can write that,
${n}''=n\dfrac{v-{{v}_{0}}}{v+{{v}_{s}}}$
Substituting the values in it will give,
${n}''=1088\times \dfrac{340-20}{340+20}=1088\times \dfrac{32}{36}=967.11Hz$
Hence the question has been answered.
Note:
The Doppler Effect is defined as phenomena in which the variation in frequency of a wave in relation to an observer who is in motion with respect to the wave source. This phenomenon is named after the Austrian physicist Christian Doppler. He is the first one who completely explained this phenomenon in 1842.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

