A source of sound is moving with constant velocity of $30 \mathrm{~m} / \mathrm{s}$ emitting a note of…
A source of sound is moving with constant velocity of $30 \mathrm{~m} / \mathrm{s}$ emitting a note of
frequency $256 \mathrm{~Hz}$. The ratio of frequencies observed by a stationary observer
while the source is approaching him and after it crosses him is
[speed of sound in air $=330 \mathrm{~m} / \mathrm{s}]$
$8: 9$
$9: 8$
$5: 6$
$6: 5$
Solution
A source of sound is moving with constant velocity of $30$ mis emitting a note of frequency $256 \mathrm{~Hz}$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him is $6: 5$.
$V_{S}=30 \mathrm{~m} / \mathrm{s} \quad\mathrm{n}_{0}=256 \mathrm{~Hz}$
$n_{1}=n_{0} \frac{V}{V-V_{s}}$
$n_{2}=n_{0} \frac{V}{V+V_{s}}$
$\therefore \frac{n_{1}}{n_{2}}=\frac{V+V_{s}}{V-V_{s}}=\frac{360}{300}=\frac{6}{5}$