Sound Waves Concept Page - 12

Example

Use the formula for observed frequency for a moving source and a moving observer

Two aeroplanes A and B are moving away from one another with a speed of 720 kmph. The frequency of the whistle emitted by A is 1100 Hz. Then find the apparent frequency of the whistle as heard by the passenger of the aeroplane B.
(velocity of sound in air is 350 Ms−1${}^{-1}$ )Given : V0=720km/h=720×10003600=200m/s$V_0=720km/h=720\times \frac{1000}{3600}=200m/s$
vs=−720km/h=200m/s$v_s=-720km/h=200m/s$
v=350m/s$v=350m/s$
δ=1100hz$\delta =1100hz$
By Doppler effect formula
δ1=δ(V−V0V+Vs)${\delta }_1=\delta \left({\displaystyle \frac{V-V_0}{V+V_s}}\right)$(source and listener are moving away from each other)
=1100(350−200350+200)$=1100\left({\displaystyle \frac{350-200}{350+200}}\right)$
=1100×150550$={\displaystyle \frac{1100\times 150}{550}}$
=300Hz

Example

Find the change in observed wavelength in case of Doppler effect

Example: A star is receding away from earth with a velocity of 105 ms−1.${10}^5{\text{ms}}^{-1}.$ If the wavelength of its spectral line is 5700 Ao$5700{\text{A}}^o$ then find the Doppler shift.

Solution:
As,we know that Doppler effect in light, =△λλ=△ff=vc$,={\displaystyle \frac{\mathrm{△}\lambda }{\lambda }}={\displaystyle \frac{\mathrm{△}f}{f}}={\displaystyle \frac{v}{c}}$
where, △λ=wavelength shift$\mathrm{△}\lambda =\text{wavelength shift}$
△f=frequency shift$\mathrm{△}f=\text{frequency shift}$
v=speed of object$v=\text{speed of object}$
c=speed of light=3×108 m/s$c=\text{speed of light}=3\times {10}^8\text{m/s}$
 △λλ=vc⇒△λ=λvc${\displaystyle \frac{\mathrm{△}\lambda }{\lambda }}={\displaystyle \frac{v}{c}}\Rightarrow \mathrm{△}\lambda ={\displaystyle \frac{\lambda v}{c}}$
△λ=5700×1053×108=1.9$\mathrm{△}\lambda ={\displaystyle \frac{5700\times {10}^5}{3\times {10}^8}}=1.9$

Result

Applications of Doppler's Effect

Applications of Doppler's effect:
1. Sirens: The siren on a passing emergency vehicle will start out higher than its stationary pitch, slide down as it passes, and continue lower than its stationary pitch as it recedes from the observer.

2. Astronomy: The Doppler effect for electromagnetic waves such as light is of great use in astronomy and results in either a so-called redshift or blueshift. It has been used to measure the speed at which stars and galaxies are approaching or receding from us; that is, their radial velocities. This may be used to detect if an apparently single star is, in reality, a close binary, to measure the rotational speed of stars and galaxies, or to detect exoplanets.

3. Other than these Doppler's effect is being used in Radar, Medical imaging and blood flow measurement, Flow measurement, Velocity profile measurement, Satellite communication, Audio, Vibration measurement, Developmental biology etc.  

Definition

State the limitations of Doppler Effect

Limitations of Doppler effect in sound are as follows:

• The Velocity of source of sound must be less than that of the Velocity of sound i.e. Vs<V$V_s<V$.
• The Velocity of observer must be less than the velocity of sound i.e. Vo<V$V_o<V$.
• If the velocity of sound of source is greater than that of Velocity of sound then due to shock waves the wave front gets distorted, consequently the change in frequency will not be observed by the observer.

Example

Solve problems in which motion of observer and source are not along the same line

A source of sound with natural frequency n moves uniformly along a straight line separated from a stationary observer by a distance of 500 metres. The velocity of source is equal to half the velocity of sound in air. Find the apparent frequency heard by observer at the instant when the source gets closest to him.Solution : Here the frequency received by the observer at the instant the source is closest to him is required. This may be calculated by noting the instant of emission of sound which is different from the instant it is received. At the instant when it is closest to observer the observer will receive these waves, which source has emitted at certain previous instant when it was at some other position P1 (say). Time taken by source to travel from P1$P_1$ to P2$P_2$ i.e., t1$t_1$ is equal to  time taken by sound waves to travel the distance P1$P_1$O(i.e.,t1$t_1$)
t1=P1P2v′=ltanθv′$t_1={\displaystyle \frac{P_1{P}_2}{v^{\prime }}}={\displaystyle \frac{l\tan \theta }{v^{\prime }}}$ where v' is the velocity of  source
t2=P1Ov=lcosθv$t_2={\displaystyle \frac{P_{1}O}{v}}={\displaystyle \frac{{\displaystyle \frac{l}{\cos \theta }}}{v}}$ where v is the velocity of sound
t1=t2$t_1=t_2$
⇒ltanθv′=lcosθv$\Rightarrow {\displaystyle \frac{l\tan \theta }{v^{\prime }}}={\displaystyle \frac{{\displaystyle \frac{l}{\cos \theta }}}{v}}$
⇒v′=vsinθ$\Rightarrow v^{\prime }=v\sin \theta$
Velocity of wave in  the direction of the source: v′cos(90−θ)=v′sinθ$v^{\prime }\cos (90-\theta )=v^{\prime }\sin \theta$
Apparent frequency n′=nvv−v′sinθ=nvv−vsinθsinθ=n1−sin2θ=n1−(v′v)2=n1−(12)2=4n3$n^{\prime }={\displaystyle \frac{nv}{v-v^{\prime }\sin \theta }}={\displaystyle \frac{nv}{v-v\sin \theta \sin \theta }}={\displaystyle \frac{n}{1-{\sin }^2\theta }}={\displaystyle \frac{n}{1-({\displaystyle \frac{v^{\prime }}{v}}{)}^2}}={\displaystyle \frac{n}{1-({\displaystyle \frac{1}{2}}{)}^2}}={\displaystyle \frac{4n}{3}}$

Example

Solve problems on Doppler effect that involve reflection of sound from a fixed surface

Example: A racing car moving towards a cliff sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound. lf V$V$ is the velocity of sound, what is the velocity of the car?
Solution:
let νc${\nu }_c$ is the apparent freq at cliff and νs${\nu }_s$ is the frequency by the sound of the car, let vs$v_s$ is the velocity of the car and V$V$  is the velocity of the sound in the air. Then from the Doppler effect,
νc=VV−vsνs${\nu }_c={\displaystyle \frac{V}{V-v_s}}{\nu }_s$,
Say the apparent frequency of the reflected sound by the cliff is ν′c${\nu }_c^{\prime }$, then
ν′c=V+vsVνc=V+vsV−vsνs${\nu }_c^{\prime }={\displaystyle \frac{V+v_s}{V}}{\nu }_c={\displaystyle \frac{V+v_s}{V-v_s}}{\nu }_s$,
But we are given, ν′c=2νs${\nu }_c^{\prime }=2{\nu }_s$
That will give us,
V+vsV−vs=2${\displaystyle \frac{V+v_s}{V-v_s}}=2$,
By Componendo-Dividendo
vs=V/3$v_s=V/3$

Example

Solve problems on Doppler effect where acceleration is involved

Example: A body emitting sound of frequency 350 Hz$350Hz$ is dropped  from a balloon rising vertically upwards with constant velocity 5 m/s$5m/s$. The frequency of sound as felt by the observer in the balloon 2 s$2s$ after the release is (Velocity of sound in air is 335 m/s$335m/s$; acceleration due to gravity is 10 ms−2$10m{s}^{-2}$).

Solution:

Velocity of observer.  vo=5ms−1$v_o=5m{s}^{-1}$
Velocity of sound  v3=gt=10(2)=20ms−1$v_3=gt=10(2)=20m{s}^{-1}$
According to Doppler's effect:
∴frequencyf′=(v−v3v+vo)f.$\therefore frequency{f}^{\prime }=({\displaystyle \frac{v-v_3}{v+v_o}})f.$
=(335−20335+5)350$=({\displaystyle \frac{335-20}{335+5}})350$
=324.26Hz.$=324.26Hz.$

Example

Solve problems on Doppler effect where non-linear motion is involved

 When a source of sound, or a listener or both are in motion relative to the air, the pitch of the sound, as heard by the listener, is in general not the same as when source and listener are at rest. The most common example is the sudden drop in pitch of sound from an automobile horn as one meets and passes a car proceeding in the opposite direction. This phenomenon is called the Doppler effect.
Let V2$V_2$ and V3$V_3$ represent the velocity of a listener and a source relative to the air. We shall consider only the special case in which the velocities lie along the line joining listener and source. Since these velocities may be in the same or opposite direction, and the listener may be either ahead or behind the source, a convention of signs is required. We shall take the positive direction of V2$V_2$ and V3$V_3$ as that from the position of the listener towards the position of the source. The speed of propagation of sound waves, C1$C_1$ will always be considered positive. We consider first listener L$L$ moving with velocity V2$V_2$ towards a stationary source S$S$.
The source emits a wave with frequency fs$f_s$ and wave length λ=c/fs${\displaystyle \lambda =c/f_s}$.
 The figure shows several waves crests, separated by equal distance λ$\lambda$.
The wave approaching the moving listener  have a speed of propagation relative to him of (c+V1)${\displaystyle (c+V_1)}$.
Thus, the frequency fL$f_L$ with which the listener encounter wave crests, i.e, the frequency he hears, is:
fL=C+VLλ=C+VLC/fs=fs(C+VL)C${\displaystyle f_L={\displaystyle \frac{C+V_L}{\lambda }}={\displaystyle \frac{C+V_L}{C/f_s}}={\displaystyle \frac{f_s(C+V_L)}{C}}}$
Example:A car is moving towards a vertical wall with 10ms−1${\displaystyle 10m{s}^{-1}}$. The driver blows a horn of frequency 400 Hz$400Hz$. The frequency heard by the drivers is ____.
Vsound=340ms−1${\displaystyle V_{sound}=340m{s}^{-1}}$

Solution: 
Frequency heard by wall where car is source and wall is object
fwall=fd1VSVS−VC$f_{wall}=f_{d1}{\displaystyle \frac{V_S}{V_S-V_C}}$  (i)
Now the reflected sound heard by driver in this case where wall is source and car is object
fd2=fwallVS+VCVS$f_{d2}=f_{wall}{\displaystyle \frac{V_S+V_C}{V_S}}$  (ii)
Using equation (i) and(ii)
fd2=fd1VS+VCVS−VC$f_{d2}=f_{d1}{\displaystyle \frac{V_S+V_C}{V_S-V_C}}$
fd2=400340+10340−10=400350330=424Hz$f_{d2}=400{\displaystyle \frac{340+10}{340-10}}=400{\displaystyle \frac{350}{330}}=424Hz$

Definition

Resultant of two interfering sound waves

Wave interference is the phenomenon that occurs when two waves meet while traveling along the same medium. Let there be two harmonic waves with angular frequencies ω1${\omega }_1$ and ω2${\omega }_2$ interfering at x=0 and having equal amplitudes and phase angles (ϕ=π2$\varphi ={\displaystyle \frac{\pi }{2}}$ each).Then the equation for the waves can be written as :
s1=acosω1t$s_1=acos{\omega }_{1}t$
s2=acosω2t$s_2=acos{\omega }_{2}t$
Then resultant wave:
s=s1+s2=a(cosω1t+cosω2t)$s=s_1+s_2=a(cos{\omega }_{1}t+cos{\omega }_{2}t)$
=2acos(ω1−ω2)t2cos(ω1+ω2)t2$=2acos{\displaystyle \frac{({\omega }_1-{\omega }_2)t}{2}}cos{\displaystyle \frac{({\omega }_1+{\omega }_2)t}{2}}$

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10
Page 11  Page 12  Page 13  Page 14  Page 15