Waves Practice Questions - DPP for JEE

 Page 1


1. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx).
Here ? = b
 take, 
? Velocity of the wave = 
= a 2pb cos 2p (bt – cx) = a? cos (?t – kx)
Maximum particle velocity  = a?= a2pb = 2p ab
given this is    i.e.  or 
2. (b) Speed of pulse at a distance x
from bottom, .
While traveling from mid point  
to the top, frequency remains
unchanged.
? 
3. (a)
Now beat frequency = f
1
 – f
2
= 
Page 2


1. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx).
Here ? = b
 take, 
? Velocity of the wave = 
= a 2pb cos 2p (bt – cx) = a? cos (?t – kx)
Maximum particle velocity  = a?= a2pb = 2p ab
given this is    i.e.  or 
2. (b) Speed of pulse at a distance x
from bottom, .
While traveling from mid point  
to the top, frequency remains
unchanged.
? 
3. (a)
Now beat frequency = f
1
 – f
2
= 
= 
= 2  
4. (a) Here, original frequency of sound, f
0
 = 100 Hz
Speed of source V
s
 = 19.4 cos 60° = 9.7
From Doppler’s formula
f = f
0
f = 100
f = 100
f = 100 = 103Hz
Apparent frequency f = 103 Hz
5. (a) For fundamental mode,
f = 
Taking logarithm on both sides, we get
Page 3


1. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx).
Here ? = b
 take, 
? Velocity of the wave = 
= a 2pb cos 2p (bt – cx) = a? cos (?t – kx)
Maximum particle velocity  = a?= a2pb = 2p ab
given this is    i.e.  or 
2. (b) Speed of pulse at a distance x
from bottom, .
While traveling from mid point  
to the top, frequency remains
unchanged.
? 
3. (a)
Now beat frequency = f
1
 – f
2
= 
= 
= 2  
4. (a) Here, original frequency of sound, f
0
 = 100 Hz
Speed of source V
s
 = 19.4 cos 60° = 9.7
From Doppler’s formula
f = f
0
f = 100
f = 100
f = 100 = 103Hz
Apparent frequency f = 103 Hz
5. (a) For fundamental mode,
f = 
Taking logarithm on both sides, we get
log f = 
= 
or log f = log 
Differentiating both sides, we get
 (as  and µ are constants)
Here df = 6
f = 600 Hz
 = 0.02
6. (b) Frequency received by listener from the rear source,
 
Frequency received by listener from the front source,
No. of beats = n’’ – n’
=  = 
7. (c) Given : Wavelength (?) = 5000 Å
velocity of star (v) = 1.5 × 10
6
 m/s.
We know that wavelength of the approaching star (?’) = 
or, 
or,  . Therefore,
Page 4


1. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx).
Here ? = b
 take, 
? Velocity of the wave = 
= a 2pb cos 2p (bt – cx) = a? cos (?t – kx)
Maximum particle velocity  = a?= a2pb = 2p ab
given this is    i.e.  or 
2. (b) Speed of pulse at a distance x
from bottom, .
While traveling from mid point  
to the top, frequency remains
unchanged.
? 
3. (a)
Now beat frequency = f
1
 – f
2
= 
= 
= 2  
4. (a) Here, original frequency of sound, f
0
 = 100 Hz
Speed of source V
s
 = 19.4 cos 60° = 9.7
From Doppler’s formula
f = f
0
f = 100
f = 100
f = 100 = 103Hz
Apparent frequency f = 103 Hz
5. (a) For fundamental mode,
f = 
Taking logarithm on both sides, we get
log f = 
= 
or log f = log 
Differentiating both sides, we get
 (as  and µ are constants)
Here df = 6
f = 600 Hz
 = 0.02
6. (b) Frequency received by listener from the rear source,
 
Frequency received by listener from the front source,
No. of beats = n’’ – n’
=  = 
7. (c) Given : Wavelength (?) = 5000 Å
velocity of star (v) = 1.5 × 10
6
 m/s.
We know that wavelength of the approaching star (?’) = 
or, 
or,  . Therefore,
[where ?? = Change in the wavelength]
8. (a) In air : T = mg = ?Vg
? f = ... (i)
In water : T = mg – upthrust
= V?g – 
? f ‘ =  = 
   
= 300 Hz
9. (a) Given wave equation is y(x,t)
=
= 
= 
= 
It is a function of type y = f (x + vt)
? Speed of wave = 
10. (b) Given 
v
t
 = v
f
B
 = 2150 Hz
Page 5


1. (a) Equation of the harmonic progressive wave given by :
y = a sin 2p (bt – cx).
Here ? = b
 take, 
? Velocity of the wave = 
= a 2pb cos 2p (bt – cx) = a? cos (?t – kx)
Maximum particle velocity  = a?= a2pb = 2p ab
given this is    i.e.  or 
2. (b) Speed of pulse at a distance x
from bottom, .
While traveling from mid point  
to the top, frequency remains
unchanged.
? 
3. (a)
Now beat frequency = f
1
 – f
2
= 
= 
= 2  
4. (a) Here, original frequency of sound, f
0
 = 100 Hz
Speed of source V
s
 = 19.4 cos 60° = 9.7
From Doppler’s formula
f = f
0
f = 100
f = 100
f = 100 = 103Hz
Apparent frequency f = 103 Hz
5. (a) For fundamental mode,
f = 
Taking logarithm on both sides, we get
log f = 
= 
or log f = log 
Differentiating both sides, we get
 (as  and µ are constants)
Here df = 6
f = 600 Hz
 = 0.02
6. (b) Frequency received by listener from the rear source,
 
Frequency received by listener from the front source,
No. of beats = n’’ – n’
=  = 
7. (c) Given : Wavelength (?) = 5000 Å
velocity of star (v) = 1.5 × 10
6
 m/s.
We know that wavelength of the approaching star (?’) = 
or, 
or,  . Therefore,
[where ?? = Change in the wavelength]
8. (a) In air : T = mg = ?Vg
? f = ... (i)
In water : T = mg – upthrust
= V?g – 
? f ‘ =  = 
   
= 300 Hz
9. (a) Given wave equation is y(x,t)
=
= 
= 
= 
It is a function of type y = f (x + vt)
? Speed of wave = 
10. (b) Given 
v
t
 = v
f
B
 = 2150 Hz
Reflected wave frequency received by A, 
Applying doppler’s effect of sound,
here, 
= 
v
t
 = 55.8372 m/s
Now, for the reflected wave,
?
= 
= 
11. (c)
The wave which reaches wall f
1
 is reflected.
The reflected frequency is f
1
 as the wall is at rest.
12. (d)
But