When a source moves away from a stationary observer, the frequency is $$ 6/7 $$ times the original frequency. Given: speed of sound = $$ 330\,m/s $$ . The speed of the source is
A
$$ 40\,m/s $$
B
$$ 55 \,m/s $$
C
$$ 330\,m/s $$
D
$$ 165 \,m/s $$
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Solution
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Correct option is B. $$ 55 \,m/s $$ $$ \nu' = \dfrac{\nu}{\nu + \nu,}\nu $$ or $$ \dfrac{6}{7} \nu = \dfrac{330}{330 + \nu,}\nu $$ or $$ 6 \times 330 + 6 \nu, = 7 \times 330 $$ or $$ 6 \nu, = 330 $$ or $$ \nu, = 55\,m/s $$
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