Answer
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Hint: Here, we have to know a little bit about the sound wave. A sound wave is a pressure disturbance that travels through a medium by means of particle to particle interaction. Since, the speed of a wave is defined as the distance at the point on a wave travels per unit of time. By using the formula: \[f' = 100k + \dfrac{{25}}{{100}} \times 100k\] to get the desired answer.
Complete answer:
Let’s take the source of the sound as VsThe observer is in rest position. f’= apparent frequency , then we take fo= actual frequency and hence $f^o=$ \[100k\]
\[
f’= 100k + \dfrac{{25}}{{100}} \times 100k \\
f' = 125k \\
\]
When source is moving towards the observer:
\[
f' = f^o[\dfrac{v}{{v - {v_s}}}] \\
125k = 100k[\dfrac{v}{{v - {v_s}}}] \\
5v - 5{v_s} = 4v \\
5{v_s} = 5v - 4v \\
5{v_s} = v \\
\therefore {v_s} = \dfrac{v}{5} \\
\]
Hence, by the above procedure we get \[\dfrac{v}{5}\] is the correct answer.
Note: It must be noted that, in air the speed of a sound wave depends upon the properties of air mostly the temperature and to a lesser degree the humidity. In a less dense material than a more dense material, a sound wave will travel faster.
Complete answer:
Let’s take the source of the sound as VsThe observer is in rest position. f’= apparent frequency , then we take fo= actual frequency and hence $f^o=$ \[100k\]
\[
f’= 100k + \dfrac{{25}}{{100}} \times 100k \\
f' = 125k \\
\]
When source is moving towards the observer:
\[
f' = f^o[\dfrac{v}{{v - {v_s}}}] \\
125k = 100k[\dfrac{v}{{v - {v_s}}}] \\
5v - 5{v_s} = 4v \\
5{v_s} = 5v - 4v \\
5{v_s} = v \\
\therefore {v_s} = \dfrac{v}{5} \\
\]
Hence, by the above procedure we get \[\dfrac{v}{5}\] is the correct answer.
Note: It must be noted that, in air the speed of a sound wave depends upon the properties of air mostly the temperature and to a lesser degree the humidity. In a less dense material than a more dense material, a sound wave will travel faster.
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