JEE Main Previous Year Questions (2025): Wave Motion and Sound wave

 Page 1


JEE Main Previous Year Questions 
(2025): Wave Motion and Sound wave 
Q1: A closed organ and an open organ tube are filled by two different gases having 
same bulk modulus but different densities ?? ?? and ?? ?? , respectively. The frequency of 
?? th 
 harmonic of closed tube is identical with ?? th 
 harmonic of open tube. If the length 
of the closed tube is ???? ???? and the density ratio of the gases is ?? ?? : ?? ?? = ?? : ???? , then 
the length of the open tube is : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 
15
7
 cm 
B. 
20
9
 cm 
C. 
20
7
 cm 
D. 
15
9
 cm 
Ans: B 
Solution: 
We know, for closed pipe, 
?? ?? =
????
4 ?? , ?? = 1 , 3 , 5 , 7 
for open pipe, ?? ?? =
????
2 ?? , ?? = 1 , 2 , 3 , 4 
So, 9
th 
 harmonic of closed pipe =
9 ?? 1
4 ?? 1
 
4
th 
 harmonic of open pipe =
4 ?? 2
2 ?? 2
=
2 ?? 2
?? 2
 
?
9 ?? 1
4 ?? 1
=
2 ?? 2
?? 2
?
?? 2
?? 1
=
8
9
?? 2
?? 1
 
We know, ?? = v
?? ?? 
So, 
?? 2
?? 1
=
v
?? 1
?? 2
 (for same ?? ) 
hence, 
?? 2
?? 1
=
8
9
v
?? 1
?? 2
=
8
9
v
1
16
=
8
9
×
1
4
 
? ?? 2
=
2
9
× ?? 1
? ?? 2
=
2
9
× 10 =
20
9
 cm 
Q2: The equation of a transverse wave travelling along a string is ?? ( ?? , ?? ) =
?? . ???????? ? [ ???? × ????
- ?? ?? + ?????? ?? ] ???? , where ?? is in mm and ?? is in second. The velocity of 
the wave is : 
Page 2


JEE Main Previous Year Questions 
(2025): Wave Motion and Sound wave 
Q1: A closed organ and an open organ tube are filled by two different gases having 
same bulk modulus but different densities ?? ?? and ?? ?? , respectively. The frequency of 
?? th 
 harmonic of closed tube is identical with ?? th 
 harmonic of open tube. If the length 
of the closed tube is ???? ???? and the density ratio of the gases is ?? ?? : ?? ?? = ?? : ???? , then 
the length of the open tube is : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 
15
7
 cm 
B. 
20
9
 cm 
C. 
20
7
 cm 
D. 
15
9
 cm 
Ans: B 
Solution: 
We know, for closed pipe, 
?? ?? =
????
4 ?? , ?? = 1 , 3 , 5 , 7 
for open pipe, ?? ?? =
????
2 ?? , ?? = 1 , 2 , 3 , 4 
So, 9
th 
 harmonic of closed pipe =
9 ?? 1
4 ?? 1
 
4
th 
 harmonic of open pipe =
4 ?? 2
2 ?? 2
=
2 ?? 2
?? 2
 
?
9 ?? 1
4 ?? 1
=
2 ?? 2
?? 2
?
?? 2
?? 1
=
8
9
?? 2
?? 1
 
We know, ?? = v
?? ?? 
So, 
?? 2
?? 1
=
v
?? 1
?? 2
 (for same ?? ) 
hence, 
?? 2
?? 1
=
8
9
v
?? 1
?? 2
=
8
9
v
1
16
=
8
9
×
1
4
 
? ?? 2
=
2
9
× ?? 1
? ?? 2
=
2
9
× 10 =
20
9
 cm 
Q2: The equation of a transverse wave travelling along a string is ?? ( ?? , ?? ) =
?? . ???????? ? [ ???? × ????
- ?? ?? + ?????? ?? ] ???? , where ?? is in mm and ?? is in second. The velocity of 
the wave is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. - 60 m / s 
B. + 60 m / s 
C. + 30 m / s 
D. - 30 m / s 
Ans: D 
Solution: 
Let's analyze the wave equation step by step. 
The given wave is: 
?? ( ?? , ?? ) = 4 . 0s i n ? [ 20 × 10
- 3
?? + 600 ?? ] mm. 
First, simplify the coefficient of ?? : 
20 × 10
- 3
= 0 . 02 mm
- 1
. 
So the equation becomes: 
?? ( ?? , ?? ) = 4 . 0s i n ? ( 0 . 02 ?? + 600 ?? ) mm. 
A standard form for a travelling wave is: 
?? ( ?? , ?? ) = ?? sin ? ( ???? - ???? ) 
which represents a wave moving in the positive ?? -direction with speed ?? =
?? ?? . 
Notice that our wave equation has the form: 
sin ? ( 0 .02 ?? + 600 ?? ) 
The positive sign in front of 600 ?? means we can rewrite the phase as: 
0 . 02 ?? + 600 ?? = 0 . 02 ?? - ( - 600 ?? ), 
which indicates that the angular frequency ?? in the standard form is effectively -600 . 
The velocity ?? of a wave is determined from the phase (for a constant phase, ?? = constant): 
???? + ???? = constant. 
Differentiating with respect to ?? : 
?? ????
????
+ ?? = 0, 
which gives: 
????
????
= -
?? ?? . 
Substituting the values: 
?? = 0 . 02 mm
- 1
 
?? = 600 s
- 1
 
We have: 
?? = -
600
0 . 02
= - 300 00 mm / s. 
Convert the velocity from mm / s to m / s : 
- 30000 mm / s = - 30 m / s. 
Thus, the velocity of the wave is - 30 m / s. 
The correct answer is Option D. 
Q3: Given below are two statements : one is labelled as Assertion A and the other is 
labelled as Reason R 
Assertion A: A sound wave has higher speed in solids than gases. 
Page 3


JEE Main Previous Year Questions 
(2025): Wave Motion and Sound wave 
Q1: A closed organ and an open organ tube are filled by two different gases having 
same bulk modulus but different densities ?? ?? and ?? ?? , respectively. The frequency of 
?? th 
 harmonic of closed tube is identical with ?? th 
 harmonic of open tube. If the length 
of the closed tube is ???? ???? and the density ratio of the gases is ?? ?? : ?? ?? = ?? : ???? , then 
the length of the open tube is : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 
15
7
 cm 
B. 
20
9
 cm 
C. 
20
7
 cm 
D. 
15
9
 cm 
Ans: B 
Solution: 
We know, for closed pipe, 
?? ?? =
????
4 ?? , ?? = 1 , 3 , 5 , 7 
for open pipe, ?? ?? =
????
2 ?? , ?? = 1 , 2 , 3 , 4 
So, 9
th 
 harmonic of closed pipe =
9 ?? 1
4 ?? 1
 
4
th 
 harmonic of open pipe =
4 ?? 2
2 ?? 2
=
2 ?? 2
?? 2
 
?
9 ?? 1
4 ?? 1
=
2 ?? 2
?? 2
?
?? 2
?? 1
=
8
9
?? 2
?? 1
 
We know, ?? = v
?? ?? 
So, 
?? 2
?? 1
=
v
?? 1
?? 2
 (for same ?? ) 
hence, 
?? 2
?? 1
=
8
9
v
?? 1
?? 2
=
8
9
v
1
16
=
8
9
×
1
4
 
? ?? 2
=
2
9
× ?? 1
? ?? 2
=
2
9
× 10 =
20
9
 cm 
Q2: The equation of a transverse wave travelling along a string is ?? ( ?? , ?? ) =
?? . ???????? ? [ ???? × ????
- ?? ?? + ?????? ?? ] ???? , where ?? is in mm and ?? is in second. The velocity of 
the wave is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. - 60 m / s 
B. + 60 m / s 
C. + 30 m / s 
D. - 30 m / s 
Ans: D 
Solution: 
Let's analyze the wave equation step by step. 
The given wave is: 
?? ( ?? , ?? ) = 4 . 0s i n ? [ 20 × 10
- 3
?? + 600 ?? ] mm. 
First, simplify the coefficient of ?? : 
20 × 10
- 3
= 0 . 02 mm
- 1
. 
So the equation becomes: 
?? ( ?? , ?? ) = 4 . 0s i n ? ( 0 . 02 ?? + 600 ?? ) mm. 
A standard form for a travelling wave is: 
?? ( ?? , ?? ) = ?? sin ? ( ???? - ???? ) 
which represents a wave moving in the positive ?? -direction with speed ?? =
?? ?? . 
Notice that our wave equation has the form: 
sin ? ( 0 .02 ?? + 600 ?? ) 
The positive sign in front of 600 ?? means we can rewrite the phase as: 
0 . 02 ?? + 600 ?? = 0 . 02 ?? - ( - 600 ?? ), 
which indicates that the angular frequency ?? in the standard form is effectively -600 . 
The velocity ?? of a wave is determined from the phase (for a constant phase, ?? = constant): 
???? + ???? = constant. 
Differentiating with respect to ?? : 
?? ????
????
+ ?? = 0, 
which gives: 
????
????
= -
?? ?? . 
Substituting the values: 
?? = 0 . 02 mm
- 1
 
?? = 600 s
- 1
 
We have: 
?? = -
600
0 . 02
= - 300 00 mm / s. 
Convert the velocity from mm / s to m / s : 
- 30000 mm / s = - 30 m / s. 
Thus, the velocity of the wave is - 30 m / s. 
The correct answer is Option D. 
Q3: Given below are two statements : one is labelled as Assertion A and the other is 
labelled as Reason R 
Assertion A: A sound wave has higher speed in solids than gases. 
Reason R: Gases have higher value of Bulk modulus than solids. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both ?? and ?? are true and ?? is the correct explanation of ?? 
B. ?? is false but ?? is true 
C. ?? is true but ?? is false 
D. Both ?? and ?? are true but ?? is NOT the correct explanation of ?? 
Ans: C 
Solution: 
We know, speed of sound in a medium ?? = v
?? ?? where B = bulk modulus, ?? = density of the 
medium 
Since, solids and liquids are much more difficult to compress than gases so they have much 
higher values of bulk modulus. 
i.e., ?? solid 
> ?? liquid 
> ?? gas 
 
Generally solids and liquids have higher mass densities ( ?? ) than gases. But corresponding 
increase in bulk modulus is much higher. 
So, ?? solid 
> ?? liquid 
> ?? gas 
 
Hence, option C is correct. 
Q4: A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the ?? -
direction in 0.3 sec . The crest ?? is at ?? = ?? at ?? = ?????? ?? and maximum displacement of 
the wave is 2 cm . Which equation correctly represents this wave? 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. ?? = 2c os ? ( 0 . 83 ?? - 3 . 35 ?? ) cm 
B. ?? = 2s i n ? ( 0 . 83 ?? - 3 . 5t ) cm 
C. ?? = 2c os ? ( 0 . 13 ?? - 0 . 5 ?? ) cm 
D. ?? = 2c o s ? ( 3 . 35 ?? - 0 . 83t ) cm 
Ans: A 
Solution: 
To find the equation of the sinusoidal wave, we need to determine several properties of the 
wave: 
Wave Velocity (v): 
Given that the wave travels a distance of 1.2 cm in 0.3 seconds, the velocity ?? can be computed 
as: 
Page 4


JEE Main Previous Year Questions 
(2025): Wave Motion and Sound wave 
Q1: A closed organ and an open organ tube are filled by two different gases having 
same bulk modulus but different densities ?? ?? and ?? ?? , respectively. The frequency of 
?? th 
 harmonic of closed tube is identical with ?? th 
 harmonic of open tube. If the length 
of the closed tube is ???? ???? and the density ratio of the gases is ?? ?? : ?? ?? = ?? : ???? , then 
the length of the open tube is : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 
15
7
 cm 
B. 
20
9
 cm 
C. 
20
7
 cm 
D. 
15
9
 cm 
Ans: B 
Solution: 
We know, for closed pipe, 
?? ?? =
????
4 ?? , ?? = 1 , 3 , 5 , 7 
for open pipe, ?? ?? =
????
2 ?? , ?? = 1 , 2 , 3 , 4 
So, 9
th 
 harmonic of closed pipe =
9 ?? 1
4 ?? 1
 
4
th 
 harmonic of open pipe =
4 ?? 2
2 ?? 2
=
2 ?? 2
?? 2
 
?
9 ?? 1
4 ?? 1
=
2 ?? 2
?? 2
?
?? 2
?? 1
=
8
9
?? 2
?? 1
 
We know, ?? = v
?? ?? 
So, 
?? 2
?? 1
=
v
?? 1
?? 2
 (for same ?? ) 
hence, 
?? 2
?? 1
=
8
9
v
?? 1
?? 2
=
8
9
v
1
16
=
8
9
×
1
4
 
? ?? 2
=
2
9
× ?? 1
? ?? 2
=
2
9
× 10 =
20
9
 cm 
Q2: The equation of a transverse wave travelling along a string is ?? ( ?? , ?? ) =
?? . ???????? ? [ ???? × ????
- ?? ?? + ?????? ?? ] ???? , where ?? is in mm and ?? is in second. The velocity of 
the wave is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. - 60 m / s 
B. + 60 m / s 
C. + 30 m / s 
D. - 30 m / s 
Ans: D 
Solution: 
Let's analyze the wave equation step by step. 
The given wave is: 
?? ( ?? , ?? ) = 4 . 0s i n ? [ 20 × 10
- 3
?? + 600 ?? ] mm. 
First, simplify the coefficient of ?? : 
20 × 10
- 3
= 0 . 02 mm
- 1
. 
So the equation becomes: 
?? ( ?? , ?? ) = 4 . 0s i n ? ( 0 . 02 ?? + 600 ?? ) mm. 
A standard form for a travelling wave is: 
?? ( ?? , ?? ) = ?? sin ? ( ???? - ???? ) 
which represents a wave moving in the positive ?? -direction with speed ?? =
?? ?? . 
Notice that our wave equation has the form: 
sin ? ( 0 .02 ?? + 600 ?? ) 
The positive sign in front of 600 ?? means we can rewrite the phase as: 
0 . 02 ?? + 600 ?? = 0 . 02 ?? - ( - 600 ?? ), 
which indicates that the angular frequency ?? in the standard form is effectively -600 . 
The velocity ?? of a wave is determined from the phase (for a constant phase, ?? = constant): 
???? + ???? = constant. 
Differentiating with respect to ?? : 
?? ????
????
+ ?? = 0, 
which gives: 
????
????
= -
?? ?? . 
Substituting the values: 
?? = 0 . 02 mm
- 1
 
?? = 600 s
- 1
 
We have: 
?? = -
600
0 . 02
= - 300 00 mm / s. 
Convert the velocity from mm / s to m / s : 
- 30000 mm / s = - 30 m / s. 
Thus, the velocity of the wave is - 30 m / s. 
The correct answer is Option D. 
Q3: Given below are two statements : one is labelled as Assertion A and the other is 
labelled as Reason R 
Assertion A: A sound wave has higher speed in solids than gases. 
Reason R: Gases have higher value of Bulk modulus than solids. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both ?? and ?? are true and ?? is the correct explanation of ?? 
B. ?? is false but ?? is true 
C. ?? is true but ?? is false 
D. Both ?? and ?? are true but ?? is NOT the correct explanation of ?? 
Ans: C 
Solution: 
We know, speed of sound in a medium ?? = v
?? ?? where B = bulk modulus, ?? = density of the 
medium 
Since, solids and liquids are much more difficult to compress than gases so they have much 
higher values of bulk modulus. 
i.e., ?? solid 
> ?? liquid 
> ?? gas 
 
Generally solids and liquids have higher mass densities ( ?? ) than gases. But corresponding 
increase in bulk modulus is much higher. 
So, ?? solid 
> ?? liquid 
> ?? gas 
 
Hence, option C is correct. 
Q4: A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the ?? -
direction in 0.3 sec . The crest ?? is at ?? = ?? at ?? = ?????? ?? and maximum displacement of 
the wave is 2 cm . Which equation correctly represents this wave? 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. ?? = 2c os ? ( 0 . 83 ?? - 3 . 35 ?? ) cm 
B. ?? = 2s i n ? ( 0 . 83 ?? - 3 . 5t ) cm 
C. ?? = 2c os ? ( 0 . 13 ?? - 0 . 5 ?? ) cm 
D. ?? = 2c o s ? ( 3 . 35 ?? - 0 . 83t ) cm 
Ans: A 
Solution: 
To find the equation of the sinusoidal wave, we need to determine several properties of the 
wave: 
Wave Velocity (v): 
Given that the wave travels a distance of 1.2 cm in 0.3 seconds, the velocity ?? can be computed 
as: 
?? =
 distance 
 time 
=
1 . 2 cm
0 . 3 s
= 4 cm / s 
Wave Number (k): 
The wave number ?? is calculated using the wavelength ?? = 7 . 5 cm : 
?? =
2 ?? ?? =
2 ?? 7 . 5
=
4 ?? 15
˜ 0 . 83 
Angular Frequency ( ?? ): 
Angular frequency ?? is related to the wave number and velocity by the equation ?? =
?? ?? , thus: 
?? = ???? = 4 ×
4 ?? 15
=
16 ?? 15
˜ 3 . 35 
Wave Equation: 
The general form for a sinusoidal wave traveling in the positive x -direction is: 
?? = ?? c os ? ( ???? - ???? ) 
Given the maximum displacement (amplitude ?? ) of the wave is 2 cm , the equation becomes: 
?? = 2c os ? ( 0 . 83 ?? - 3 . 35 ?? ) cm 
This equation accurately describes the sinusoidal wave given the provided parameters. 
Q5: In the resonance experiment, two air columns (closed at one end) of 100 cm and 
120 cm long, give 15 beats per second when each one is sounding in the respective 
fundamental modes. The velocity of sound in the air column is: 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 370 m / s 
B. 340 m / s 
C. 335 m / s 
D. 360 m / s 
Ans: D 
Solution: 
The fundamental frequency for a closed (organ) pipe can be expressed as: 
?? =
?? 4 l
 
For the first air column, with length l
1
, the frequency ?? 1
 is: 
?? 1
=
?? 4 l
1
 
For the second air column, with length l
2
, the frequency ?? 2
 is: 
?? 2
=
?? 4 l
2
 
The beat frequency, which is the difference in these two frequencies ( ?? 1
- ?? 2
 ), is given as 15 
beats per second: 
Beat = ?? 1
- ?? 2
=
?? 4
(
1
l
1
-
1
l
2
) 
Substitute the given lengths into the formula: 
15 =
?? 4
(
1
1
-
1
1 . 2
) 
Simplify the equation: 
Page 5


JEE Main Previous Year Questions 
(2025): Wave Motion and Sound wave 
Q1: A closed organ and an open organ tube are filled by two different gases having 
same bulk modulus but different densities ?? ?? and ?? ?? , respectively. The frequency of 
?? th 
 harmonic of closed tube is identical with ?? th 
 harmonic of open tube. If the length 
of the closed tube is ???? ???? and the density ratio of the gases is ?? ?? : ?? ?? = ?? : ???? , then 
the length of the open tube is : 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 
15
7
 cm 
B. 
20
9
 cm 
C. 
20
7
 cm 
D. 
15
9
 cm 
Ans: B 
Solution: 
We know, for closed pipe, 
?? ?? =
????
4 ?? , ?? = 1 , 3 , 5 , 7 
for open pipe, ?? ?? =
????
2 ?? , ?? = 1 , 2 , 3 , 4 
So, 9
th 
 harmonic of closed pipe =
9 ?? 1
4 ?? 1
 
4
th 
 harmonic of open pipe =
4 ?? 2
2 ?? 2
=
2 ?? 2
?? 2
 
?
9 ?? 1
4 ?? 1
=
2 ?? 2
?? 2
?
?? 2
?? 1
=
8
9
?? 2
?? 1
 
We know, ?? = v
?? ?? 
So, 
?? 2
?? 1
=
v
?? 1
?? 2
 (for same ?? ) 
hence, 
?? 2
?? 1
=
8
9
v
?? 1
?? 2
=
8
9
v
1
16
=
8
9
×
1
4
 
? ?? 2
=
2
9
× ?? 1
? ?? 2
=
2
9
× 10 =
20
9
 cm 
Q2: The equation of a transverse wave travelling along a string is ?? ( ?? , ?? ) =
?? . ???????? ? [ ???? × ????
- ?? ?? + ?????? ?? ] ???? , where ?? is in mm and ?? is in second. The velocity of 
the wave is : 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. - 60 m / s 
B. + 60 m / s 
C. + 30 m / s 
D. - 30 m / s 
Ans: D 
Solution: 
Let's analyze the wave equation step by step. 
The given wave is: 
?? ( ?? , ?? ) = 4 . 0s i n ? [ 20 × 10
- 3
?? + 600 ?? ] mm. 
First, simplify the coefficient of ?? : 
20 × 10
- 3
= 0 . 02 mm
- 1
. 
So the equation becomes: 
?? ( ?? , ?? ) = 4 . 0s i n ? ( 0 . 02 ?? + 600 ?? ) mm. 
A standard form for a travelling wave is: 
?? ( ?? , ?? ) = ?? sin ? ( ???? - ???? ) 
which represents a wave moving in the positive ?? -direction with speed ?? =
?? ?? . 
Notice that our wave equation has the form: 
sin ? ( 0 .02 ?? + 600 ?? ) 
The positive sign in front of 600 ?? means we can rewrite the phase as: 
0 . 02 ?? + 600 ?? = 0 . 02 ?? - ( - 600 ?? ), 
which indicates that the angular frequency ?? in the standard form is effectively -600 . 
The velocity ?? of a wave is determined from the phase (for a constant phase, ?? = constant): 
???? + ???? = constant. 
Differentiating with respect to ?? : 
?? ????
????
+ ?? = 0, 
which gives: 
????
????
= -
?? ?? . 
Substituting the values: 
?? = 0 . 02 mm
- 1
 
?? = 600 s
- 1
 
We have: 
?? = -
600
0 . 02
= - 300 00 mm / s. 
Convert the velocity from mm / s to m / s : 
- 30000 mm / s = - 30 m / s. 
Thus, the velocity of the wave is - 30 m / s. 
The correct answer is Option D. 
Q3: Given below are two statements : one is labelled as Assertion A and the other is 
labelled as Reason R 
Assertion A: A sound wave has higher speed in solids than gases. 
Reason R: Gases have higher value of Bulk modulus than solids. 
In the light of the above statements, choose the correct answer from the options 
given below: 
JEE Main 2025 (Online) 28th January Morning Shift 
Options: 
A. Both ?? and ?? are true and ?? is the correct explanation of ?? 
B. ?? is false but ?? is true 
C. ?? is true but ?? is false 
D. Both ?? and ?? are true but ?? is NOT the correct explanation of ?? 
Ans: C 
Solution: 
We know, speed of sound in a medium ?? = v
?? ?? where B = bulk modulus, ?? = density of the 
medium 
Since, solids and liquids are much more difficult to compress than gases so they have much 
higher values of bulk modulus. 
i.e., ?? solid 
> ?? liquid 
> ?? gas 
 
Generally solids and liquids have higher mass densities ( ?? ) than gases. But corresponding 
increase in bulk modulus is much higher. 
So, ?? solid 
> ?? liquid 
> ?? gas 
 
Hence, option C is correct. 
Q4: A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the ?? -
direction in 0.3 sec . The crest ?? is at ?? = ?? at ?? = ?????? ?? and maximum displacement of 
the wave is 2 cm . Which equation correctly represents this wave? 
JEE Main 2025 (Online) 2nd April Evening Shift 
Options: 
A. ?? = 2c os ? ( 0 . 83 ?? - 3 . 35 ?? ) cm 
B. ?? = 2s i n ? ( 0 . 83 ?? - 3 . 5t ) cm 
C. ?? = 2c os ? ( 0 . 13 ?? - 0 . 5 ?? ) cm 
D. ?? = 2c o s ? ( 3 . 35 ?? - 0 . 83t ) cm 
Ans: A 
Solution: 
To find the equation of the sinusoidal wave, we need to determine several properties of the 
wave: 
Wave Velocity (v): 
Given that the wave travels a distance of 1.2 cm in 0.3 seconds, the velocity ?? can be computed 
as: 
?? =
 distance 
 time 
=
1 . 2 cm
0 . 3 s
= 4 cm / s 
Wave Number (k): 
The wave number ?? is calculated using the wavelength ?? = 7 . 5 cm : 
?? =
2 ?? ?? =
2 ?? 7 . 5
=
4 ?? 15
˜ 0 . 83 
Angular Frequency ( ?? ): 
Angular frequency ?? is related to the wave number and velocity by the equation ?? =
?? ?? , thus: 
?? = ???? = 4 ×
4 ?? 15
=
16 ?? 15
˜ 3 . 35 
Wave Equation: 
The general form for a sinusoidal wave traveling in the positive x -direction is: 
?? = ?? c os ? ( ???? - ???? ) 
Given the maximum displacement (amplitude ?? ) of the wave is 2 cm , the equation becomes: 
?? = 2c os ? ( 0 . 83 ?? - 3 . 35 ?? ) cm 
This equation accurately describes the sinusoidal wave given the provided parameters. 
Q5: In the resonance experiment, two air columns (closed at one end) of 100 cm and 
120 cm long, give 15 beats per second when each one is sounding in the respective 
fundamental modes. The velocity of sound in the air column is: 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 370 m / s 
B. 340 m / s 
C. 335 m / s 
D. 360 m / s 
Ans: D 
Solution: 
The fundamental frequency for a closed (organ) pipe can be expressed as: 
?? =
?? 4 l
 
For the first air column, with length l
1
, the frequency ?? 1
 is: 
?? 1
=
?? 4 l
1
 
For the second air column, with length l
2
, the frequency ?? 2
 is: 
?? 2
=
?? 4 l
2
 
The beat frequency, which is the difference in these two frequencies ( ?? 1
- ?? 2
 ), is given as 15 
beats per second: 
Beat = ?? 1
- ?? 2
=
?? 4
(
1
l
1
-
1
l
2
) 
Substitute the given lengths into the formula: 
15 =
?? 4
(
1
1
-
1
1 . 2
) 
Simplify the equation: 
15 =
?? 4
(
0 . 2
1 . 2
) 
Solve for ?? : 
?? =
15 × 4 × 1 . 2
0 . 2
= 60 × 6 = 360 m / s 
Thus, the velocity of sound in the air column is ?????? ?? / ?? . 
Q6: In an experiment with a closed organ pipe, it is filled with water by (
?? ?? ) th of its 
volume. The frequency of the fundamental note will change by 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 20% 
B. 25% 
C. - 20% 
D. - 25% 
Ans: B 
Solution: 
 
?? 1
= 4 l
f
1
=
v
4 l