Consider a large plane diaphragm ‘S’ emitting sound and a detector ‘O’. The diagram shows plane wave fronts for the sound wave travelling in air towards right when source, observer and medium are at rest. A A' and BB' are fixed imaginary planes. Column-I describes about the motion of source, observer or medium and column-II describes various effects. Match them correctly. 

 

 

	Column  I		Column  II
a)	Source starts moving  towards right.	p)	Distance between any two wave fronts will increase.
b)	Air starts moving towards right.	q)	Distance between any two wave fronts will decrease.
c)	Observer and source both move towards left with same speed.	r)	The time needed by sound to move from plane A A' to BB' will increase.
d)	Source and medium (air) both move towards right with same speed.	s)	The time needed by sound to move from plane AA' to BB' will decrease.
		t)	Frequency received by observer increases.

$[a\to q,t or q,s,t;b\to p,s or p;c\to p or p,r; d\to s,t]$
When source move towards observer w.r.t. medium, distance between two consecutive wave-front decreases and there by decreasing time taken in travelling from one wave front to other and hence increasing frequency.
In case of ‘A’ source moves towards observer w.r.t. medium while in B and C source moves away from medium while in B and C source moves away from medium. 
(a) [q, t]
If A A' and BB' are considered to be fixed planes, then source moves towards right.

v-speed of sound   
d-remains same
v-remains same
$t=\frac{d}{v}$ =remains the same
(a) [q, s, t]
If A A' and BB' are considered as wave fronts if sources moves toward right.

$f\text{'}=\frac{v}{\lambda \text{'}} \lambda \text{'}=\frac{vt-v_{s}t}{ft}=\left(\frac{v-v_s}{f}\right)$

$$\begin{gathered} f\text{'}=f\left(\frac{v}{v-v_s}\right) \\ f\text{'}>f \end{gathered}$$

Since distance between $AA\text{'}$ and $BB\text{'}$ wave front decrease hence time taken by sound wave decreases.
b)  [P,S] 
     If $AA\text{'}$ and $BB\text{'}$ are considered as fixed planes. Air moves toward right.

$t\text{'}=\frac{d}{v+v_{air}}\equiv decreases$

b) [P]
  If $AA\text{'}$ and $BB\text{'}$ are considered as wave fronts distance between then increases.
 Initially $t=\frac{d}{v}=\frac{\lambda }{v}=\frac{1}{f}$ 
$x\text{'}=\frac{v+v_{air}}{f}(v+v_{air})=v_{net}$ (net velocity of sound)
$$\begin{gathered} \frac{(v+v_{air})}{f(v+air)}=t\text{'}=\frac{d\text{'}}{v+v_{air}}\equiv remainsame \\ t\text{'}=\frac{1}{f} \end{gathered}$$

c) [P]
If  If $AA\text{'}$ and $BB\text{'}$ are considered to be fixed planes and source and observer move toward left.
d-remains same  
$t=\frac{d}{v}\equiv remainsame$

c) [p,r]
If $AA\text{'}$and $BB\text{'}$are considered wave fronts, distance between them increase and hence time taken by sound increases.

d) [s,t]
If $AA\text{'}$and $BB\text{'}$ are considered fixed planes source and medium both move toward right due to motion of medium.

$$\begin{gathered} t=d/v \\ d\to remainsame \\ {v}_{net}=v+v_{air} \end{gathered}$$

(net velocity of sound increases)

$t\text{'}=\frac{d}{(v+v_{air})}=decreases$