Waves Test - 12

The reason for the Doppler effect is that when the source of the waves is moving towards the observer, each successive wave crest is emitted from a position closer to the observer than the crest of the previous wave.

F(s)/F(l) = [V+V(s)]/[V+V(l)]
V = velocity of sound = 340m/s
V(l) = velocity of listener = 0
F(l) = frequency heard by listener = ?
V(s) = velocity of source = -20m/s (because, it's source to listener)
F(s) = frequency of source = 600hz
By putting these values in the above formula and solving we get,
F(l) = 637.5 Hz.
Hence C is correct.

Given:
Frequency of radio signal n=9×10 ⁹Hz
 Frequency shift n0=3×10³Hz
 Velocity of radio signal v=3×10⁸m/s
 Frequency shift shown by reflected wave is Shift =n−n
=>n−n=(v+u/v−u)n−n
 
Shift=(2u/v−u)ⁿ
Now, putting the given values in Eq. (i), we get,
3×10³=(2u/3×10^-8−u)9×10⁹
                    
⇒3×10⁸−u=2u×9×10⁹/3×10³
=6u×10⁶
⇒6×10⁶u+u=3×10⁸
 
u(6×106+1)=3×10⁸
 
u(6000001)=3×10⁸
 
u=3×10⁸/6000001=49.999≈50m/s

In the first situation, a police car (source of sound) with a speed v_s​, is approaching a motorcycle (observer), which is moving away from a police car with a speed of v_m​. 
Therefore, apparent frequency of sound heard by motorcyclist,

n′=n_car​(​v−v_m /v−v_s ​​) ...................eq1
.
In second situation, motorcyclist (observer) is approaching a stationary siren (source), with a speed of vm​. 
Therefore, apparent frequency of sound heard by motorcyclist,

n′′=n_siren​(v+v_m/v​​) ...................eq2 ,.

Now as no beats are observed by motorcyclist, this is possible only when difference in frequencies heard by motorcyclist is zero,
i.e.   n′−n′′=0,
or    n′=n′′,
or    n_car​(​v−v_m /v−v_s ​​)=n_siren​(v+vm/v​​)
Given n′=176Hz, n′′=165Hz, v=330m/s, v_s​=22m/s,ncar​=176Hz  and  n_siren​=165Hz 
Hence, 176(v−v_m​​/330−22) =165(v+v_m/330​​),
or          (​v−v_m/ v+v_m ​​)=165​/176×308/330​=7/8 ,
or         15v_m​=v,
Speed of sound in air v=330 m/s
or         v_m​=330/15=22m/s

Sound waves require a material medium for their propagation. Therefore we say that the Doppler Effect in sound is asymmetric.

In medicine, the Doppler Effect can be used to measure the direction and speed of blood flow in arteries and veins. This is used in echocardiograms and medical ultrasonography and is an effective tool in diagnosis of vascular problems.

From Doppler's effect
1000=9500(300/300−v​)⇒v=15m/s

The Doppler Effect is observed whenever the speed of the source is moving slower than the speed of waves. But if the source actually moves at the same speed as or faster than the waves, a different phenomenon is observed. This phenomenon is known as Shock waves or Sonic Booms.

When no beats is listened at the station thus we get the relative frequencies of both the trains are the same.
Hence the relative frequency of the train B is 600 (330/300) = 660
similarly the relative frequency of train A is f(330/315) = 660
Hence we get f = 630Hz