- Home
- Standard 11
- Physics
The apparent frequency of a sound wave as heard by an observer is $10\%$ more than the actual frequency. If the velocity of sound in air is $330\, m/sec$, then
$(i)$ The source may be moving towards the observer with a velocity of $30\,ms^{-1}$
$(ii)$ The source may be moving towards the observer with a velocity of $33\,ms^{-1}$
$(iii)$ The observer may be moving towards the source with a velocity of $30\,ms^{-1}$
$(iv)$ The observer may be moving towards the source with a velocity of $33\,ms^{-1}$
$ii, iv$
$ii, iii$
$i, iv$
$iii, iv$
Solution
$f_{\mathrm{app}}=f_{\mathrm{actual}}+10 \%$ of factual
$=\mathrm{f}+\frac{10}{100} \mathrm{f}=1.1 \mathrm{f}=\mathrm{f}\left[\frac{\mathrm{v}-\mathrm{v}_{0}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right]$
$\Rightarrow \frac{v-v_{0}}{v-v_{s}}=1.1$
if $v_{0}=0 \Rightarrow v=1.1 \mathrm{v}-1.1 \mathrm{v}_{\mathrm{s}}$
$\Rightarrow v_{s}=\frac{0.1 \mathrm{v}}{1.1}=\frac{330}{11}=30 \mathrm{\,m} / \mathrm{sec}$
$\mathrm{v}_{\mathrm{s}}=30 \mathrm{\,m} / \mathrm{s}$
[source moving towards stationary observer]
Case $II \,:$ $v_{s}=0$
$\Rightarrow 1.1=\left[\frac{\mathrm{v}-\mathrm{v}_{0}}{\mathrm{v}}\right]$
$\Rightarrow 1.1 \mathrm{v}=\mathrm{v}-\mathrm{v}_{0}$
$\Rightarrow \mathrm{v}_{0}=-0.1 \mathrm{v}$
$\mathrm{v}_{0}=-33 \mathrm{\,m} / \mathrm{sec}$
(observer moving towards source)
