The difference between the apparent frequency of a source of sound as perceived by an observer during its approach and recession is $2\%$ of the natural frequency of the source. If the velocity of sound in air is $300 \,m/sec,$ the velocity of the source is ... $m/sec$ (It is given that velocity of source $<<$ velocity of sound)
Diffcult
Download our app for free and get started
(b) When the source approaches the observer
Apparent frequency $n' = \frac{v}{{v - {v_s}}}.n = n\left[ {\frac{1}{{1 - \frac{{{v_s}}}{v}}}} \right]$
=$n{\left[ {1 - \frac{{{v_s}}}{v}} \right]^{ - 1}} = n\left[ {1 + \frac{{{v_s}}}{v}} \right]$
(Neglecting higher powers because $v_S << v$)
When the source recedes the observed apparent frequency $n'' = n\left[ {1 - \frac{{{v_s}}}{v}} \right]$
Given $n' - n'' = \frac{2}{{100}}n,\,\,v = 300$$m/\sec $
$\therefore $ $\frac{2}{{100}}n = n\left[ {1 + \frac{{{v_s}}}{v}} \right] - n\left[ {1 - \frac{{{v_s}}}{v}} \right] = n\left[ {2\frac{{{v_s}}}{v}} \right]$
$ \Rightarrow $ $\frac{2}{{100}} = 2\frac{{{v_s}}}{v} \Rightarrow {v_s} = \frac{v}{{100}} = \frac{{300}}{{100}} = 3$$m/\sec $.
Apparent frequency $n' = \frac{v}{{v - {v_s}}}.n = n\left[ {\frac{1}{{1 - \frac{{{v_s}}}{v}}}} \right]$
=$n{\left[ {1 - \frac{{{v_s}}}{v}} \right]^{ - 1}} = n\left[ {1 + \frac{{{v_s}}}{v}} \right]$
(Neglecting higher powers because $v_S << v$)
When the source recedes the observed apparent frequency $n'' = n\left[ {1 - \frac{{{v_s}}}{v}} \right]$
Given $n' - n'' = \frac{2}{{100}}n,\,\,v = 300$$m/\sec $
$\therefore $ $\frac{2}{{100}}n = n\left[ {1 + \frac{{{v_s}}}{v}} \right] - n\left[ {1 - \frac{{{v_s}}}{v}} \right] = n\left[ {2\frac{{{v_s}}}{v}} \right]$
$ \Rightarrow $ $\frac{2}{{100}} = 2\frac{{{v_s}}}{v} \Rightarrow {v_s} = \frac{v}{{100}} = \frac{{300}}{{100}} = 3$$m/\sec $.
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1What will be the wave velocity, if the radar gives $54$ waves per min and wavelength of the given wave is $10\, m$ ...... $m/sec$View Solution
- 2A copper wire is held at the two ends by rigid supports. At $50^{\circ} C$ the wire is just taut, with negligible tension. If $Y=1.2 \times 10^{11} \,N / m ^2, \alpha=1.6 \times 10^{-5} /{ }^{\circ} C$ and $\rho=9.2 \times 10^3 \,kg / m ^3$, then the speed of transverse waves in this wire at $30^{\circ} C$ is .......... $m / s$View Solution
- 3The speed of sound in a gas, in which two waves of wavelength $1.0\,\, m$ and $1.02\,\, m$ produce $6$ beats per second, is approximately .... $m/s$View Solution
- 4A vehicle with a horn of frequency $n$ is moving with a velocity of $30\, m/s$ in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency $n + {n_1}$. Then (if the sound velocity in air is $300\, m/s$)View Solution
- 5The vibrations of a string of length $60\, cm$ fixed at both the ends are represented by the equation $y = 2\,\sin \,\left( {\frac{{4\pi x}}{{15}}} \right)\,\cos \,\left( {96\pi t} \right)$ where $x$ and $y$ are in $cm$. The maximum number of loops that can be formed in it isView Solution
- 6Equation of a progressive wave is given by $y = 0.2\cos \pi \left( {0.04t + 0.02x - \frac{\pi }{6}} \right)$The distance is expressed in $cm$ and time in second. What will be the minimum distance between two particles having the phase difference of $\pi /2$ ...... $cm$View Solution
- 7The frequency changes by $10\%$ as a sound source approaches a stationary observer with constant speed $v_s$. What would be the percentage change in frequency as the source recedes the observer with the same speed. ... $\%$ Given that $v_s < v$. ($v =$ speed of sound in air)View Solution
- 8Tuning fork ${F_1}$ has a frequency of $256 Hz$ and it is observed to produce $6$ beats/second with another tuning fork ${F_2}$. When ${F_2}$ is loaded with wax, it still produces $6$ beats/second with ${F_1}$. The frequency of ${F_2}$ before loading was ..... $Hz$View Solution
- 9A travelling harmonic wave is represented by the equation $y(x, t) = 10^{-3}\,sin\,(50t + 2x)$, where $x$ and $y$ are in meter and $t$ is in seconds. Which of the following is a correct statement about the wave?View Solution
- 10A stationary source emits sound waves of frequency $500\, Hz$. Two observers moving along a line passing through the source detect sound to be of frequencies $480\, Hz$ and $530\, Hz$. Their respective speeds are, in $m\,s^{-1}$ (Given speed of sound $= 300\, m/s$)View Solution