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# A source of sound is travelling towards a stationary observer. The frequency of sound heard by the observer in $25\%$ is more than that of the actual frequency if the speed of sound is V, that of the source is.

Last updated date: 24th Jul 2024
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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.

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.