An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

5 %

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20 %

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0 %

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0.5 %

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Solution

The correct option is B $20 %$
Apparent Frequency is given by:
$f = f_{0} (\frac{v_{s} + v_{0}}{v_{s}})$
$f = f_{0} [\frac{v + \frac{v}{5}}{v}]$
$f = \frac{6}{5} f_{0}$

Hence, percentage increase is
$p = ⎡ ⎣ \frac{\frac{6}{5} f_{0} - f_{0}}{f_{0}} ⎤ ⎦ \times 100$
$p = 20 %$

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An observer moves towards a stationary source of sound with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

The correct option is B 20%Apparent Frequency is given by: f=f0(vs+v0vs) f=f0[v+v5v] f=65f0 Hence, percentage increase is p=⎡⎣65f0−f0f0⎤⎦×100 p=20%

The correct option is B $20 %$
Apparent Frequency is given by:
$f = f_{0} (\frac{v_{s} + v_{0}}{v_{s}})$
$f = f_{0} [\frac{v + \frac{v}{5}}{v}]$
$f = \frac{6}{5} f_{0}$

Hence, percentage increase is
$p = ⎡ ⎣ \frac{\frac{6}{5} f_{0} - f_{0}}{f_{0}} ⎤ ⎦ \times 100$
$p = 20 %$