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Waves: JEE Main Previous Year Questions (2021-2026)
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Page 1
JEE Main Previous Year Questions (2021-2026):
Waves
(January 2026)
Q1: Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When
fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original
frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________
Hz.
Ans: 384
Sol:
Before waxing the fork, A
The beat frequency is the number of beats per second.
This means the absolute difference between the frequencies of fork A (f
A
) and fork B (f
B
) is 4 Hz
:
Since f
B
= 380 Hz, the original frequency of fork A could be either :
Case 1 : f
A
= 380 + 4 = 384 Hz
Case 2 : f
A
= 380 - 4 = 376 Hz
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to
decrease. Let the new frequency of fork A be f'A , where f'
A
< f
A
.
The new beat frequency (f'b) is :
Initially, f
A
is higher than f
B
by 4 Hz (384 - 380 = 4).
If f
A
decreases (to say 382 Hz), the gap between f
A
and f
B
decreases (382 - 380 = 2).
The original frequency of tuning fork A must have been higher than fork B to allow the beat
frequency to decrease as f
A
dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is 384.
Q2: The velocity of sound in air is doubled when the temperature is raised from 0
°
C to
a
°
C . The value of a is______.
Ans: 819
Sol:
Page 2
JEE Main Previous Year Questions (2021-2026):
Waves
(January 2026)
Q1: Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When
fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original
frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________
Hz.
Ans: 384
Sol:
Before waxing the fork, A
The beat frequency is the number of beats per second.
This means the absolute difference between the frequencies of fork A (f
A
) and fork B (f
B
) is 4 Hz
:
Since f
B
= 380 Hz, the original frequency of fork A could be either :
Case 1 : f
A
= 380 + 4 = 384 Hz
Case 2 : f
A
= 380 - 4 = 376 Hz
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to
decrease. Let the new frequency of fork A be f'A , where f'
A
< f
A
.
The new beat frequency (f'b) is :
Initially, f
A
is higher than f
B
by 4 Hz (384 - 380 = 4).
If f
A
decreases (to say 382 Hz), the gap between f
A
and f
B
decreases (382 - 380 = 2).
The original frequency of tuning fork A must have been higher than fork B to allow the beat
frequency to decrease as f
A
dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is 384.
Q2: The velocity of sound in air is doubled when the temperature is raised from 0
°
C to
a
°
C . The value of a is______.
Ans: 819
Sol:
The velocity of sound in an ideal gas depends on the absolute temperature of the medium.
The speed of sound (v) in air is given by the formula:
From this, we can see that :
where T is the absolute temperature in Kelvin (K).
Initial State :
Temperature
Velocity = v
1
Final State :
Temperature
Velocity = v
2
= 2v
1
(since the velocity is doubled).
Using the proportionality
Substituting the values :
Page 3
JEE Main Previous Year Questions (2021-2026):
Waves
(January 2026)
Q1: Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When
fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original
frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________
Hz.
Ans: 384
Sol:
Before waxing the fork, A
The beat frequency is the number of beats per second.
This means the absolute difference between the frequencies of fork A (f
A
) and fork B (f
B
) is 4 Hz
:
Since f
B
= 380 Hz, the original frequency of fork A could be either :
Case 1 : f
A
= 380 + 4 = 384 Hz
Case 2 : f
A
= 380 - 4 = 376 Hz
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to
decrease. Let the new frequency of fork A be f'A , where f'
A
< f
A
.
The new beat frequency (f'b) is :
Initially, f
A
is higher than f
B
by 4 Hz (384 - 380 = 4).
If f
A
decreases (to say 382 Hz), the gap between f
A
and f
B
decreases (382 - 380 = 2).
The original frequency of tuning fork A must have been higher than fork B to allow the beat
frequency to decrease as f
A
dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is 384.
Q2: The velocity of sound in air is doubled when the temperature is raised from 0
°
C to
a
°
C . The value of a is______.
Ans: 819
Sol:
The velocity of sound in an ideal gas depends on the absolute temperature of the medium.
The speed of sound (v) in air is given by the formula:
From this, we can see that :
where T is the absolute temperature in Kelvin (K).
Initial State :
Temperature
Velocity = v
1
Final State :
Temperature
Velocity = v
2
= 2v
1
(since the velocity is doubled).
Using the proportionality
Substituting the values :
Therefore, the value of a is 819. Hence, the correct answer is 819.
Q3: Two loudspeakers (L
1
and L
2
) are placed with a separation of 10 m, as shown in
figure. Both speakers are fed with an audio input signal of same frequency with constant
volume. A voice recorder, initially at point A , at equidistance to both loud speakers, is
moved by 25 m along the line A B while monitoring the audio signal. The measured
signal was found to undergo 10 cycles of minima and maxima during the movement. The
frequency of the input signal is ______ Hz (Speed of sound in air is 324 m / s and v5 =
2.23)
Ans: 600
Sol:
Page 4
JEE Main Previous Year Questions (2021-2026):
Waves
(January 2026)
Q1: Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When
fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original
frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________
Hz.
Ans: 384
Sol:
Before waxing the fork, A
The beat frequency is the number of beats per second.
This means the absolute difference between the frequencies of fork A (f
A
) and fork B (f
B
) is 4 Hz
:
Since f
B
= 380 Hz, the original frequency of fork A could be either :
Case 1 : f
A
= 380 + 4 = 384 Hz
Case 2 : f
A
= 380 - 4 = 376 Hz
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to
decrease. Let the new frequency of fork A be f'A , where f'
A
< f
A
.
The new beat frequency (f'b) is :
Initially, f
A
is higher than f
B
by 4 Hz (384 - 380 = 4).
If f
A
decreases (to say 382 Hz), the gap between f
A
and f
B
decreases (382 - 380 = 2).
The original frequency of tuning fork A must have been higher than fork B to allow the beat
frequency to decrease as f
A
dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is 384.
Q2: The velocity of sound in air is doubled when the temperature is raised from 0
°
C to
a
°
C . The value of a is______.
Ans: 819
Sol:
The velocity of sound in an ideal gas depends on the absolute temperature of the medium.
The speed of sound (v) in air is given by the formula:
From this, we can see that :
where T is the absolute temperature in Kelvin (K).
Initial State :
Temperature
Velocity = v
1
Final State :
Temperature
Velocity = v
2
= 2v
1
(since the velocity is doubled).
Using the proportionality
Substituting the values :
Therefore, the value of a is 819. Hence, the correct answer is 819.
Q3: Two loudspeakers (L
1
and L
2
) are placed with a separation of 10 m, as shown in
figure. Both speakers are fed with an audio input signal of same frequency with constant
volume. A voice recorder, initially at point A , at equidistance to both loud speakers, is
moved by 25 m along the line A B while monitoring the audio signal. The measured
signal was found to undergo 10 cycles of minima and maxima during the movement. The
frequency of the input signal is ______ Hz (Speed of sound in air is 324 m / s and v5 =
2.23)
Ans: 600
Sol:
When sound waves from two coherent sources (like the two loudspeakers) meet, they interfere
with each other.
Constructive Interference (Maxima) occurs when the waves are in phase.
The path difference, ? x is an integer multiple of the wavelength (?).
where ?x = n? … (where n = 0 , 1 , 2 , …)
Destructive Interference (Minima) occurs when the waves are out of phase. This happens when
the path difference is a half-integer multiple of the wavelength.
It is given that A is equidistant from both speakers (L
1 A
= L
2 A
). Therefore, the path difference at
A is zero (?x
A
= 0) . This corresponds to the central maximum (n = 0).
As the recorder moves towards B , the distance to L
2
increases more than the distance to L
1
, so
the path difference (?x) steadily increases.
The signal undergoes 10 cycles of minima and maxima. One complete cycle means moving
from one maximum to the next consecutive maximum, which corresponds to an increase in path
difference by exactly one wavelength (?).
Since it goes through 10 full cycles starting from the central maximum, the path difference at
point B must be exactly 10 wavelengths.
?x
B
= 10?
Page 5
JEE Main Previous Year Questions (2021-2026):
Waves
(January 2026)
Q1: Two tuning forks A and B are sounded together giving rise to 8 beats in 2 s. When
fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 s. If the original
frequency of tuning fork B is 380 Hz then original frequency of tuning fork A is _________
Hz.
Ans: 384
Sol:
Before waxing the fork, A
The beat frequency is the number of beats per second.
This means the absolute difference between the frequencies of fork A (f
A
) and fork B (f
B
) is 4 Hz
:
Since f
B
= 380 Hz, the original frequency of fork A could be either :
Case 1 : f
A
= 380 + 4 = 384 Hz
Case 2 : f
A
= 380 - 4 = 376 Hz
When a tuning fork is loaded with wax, its mass increases, which causes its frequency to
decrease. Let the new frequency of fork A be f'A , where f'
A
< f
A
.
The new beat frequency (f'b) is :
Initially, f
A
is higher than f
B
by 4 Hz (384 - 380 = 4).
If f
A
decreases (to say 382 Hz), the gap between f
A
and f
B
decreases (382 - 380 = 2).
The original frequency of tuning fork A must have been higher than fork B to allow the beat
frequency to decrease as f
A
dropped.
Therefore, the original frequency of tuning fork A is 384 Hz . Hence, the correct answer is 384.
Q2: The velocity of sound in air is doubled when the temperature is raised from 0
°
C to
a
°
C . The value of a is______.
Ans: 819
Sol:
The velocity of sound in an ideal gas depends on the absolute temperature of the medium.
The speed of sound (v) in air is given by the formula:
From this, we can see that :
where T is the absolute temperature in Kelvin (K).
Initial State :
Temperature
Velocity = v
1
Final State :
Temperature
Velocity = v
2
= 2v
1
(since the velocity is doubled).
Using the proportionality
Substituting the values :
Therefore, the value of a is 819. Hence, the correct answer is 819.
Q3: Two loudspeakers (L
1
and L
2
) are placed with a separation of 10 m, as shown in
figure. Both speakers are fed with an audio input signal of same frequency with constant
volume. A voice recorder, initially at point A , at equidistance to both loud speakers, is
moved by 25 m along the line A B while monitoring the audio signal. The measured
signal was found to undergo 10 cycles of minima and maxima during the movement. The
frequency of the input signal is ______ Hz (Speed of sound in air is 324 m / s and v5 =
2.23)
Ans: 600
Sol:
When sound waves from two coherent sources (like the two loudspeakers) meet, they interfere
with each other.
Constructive Interference (Maxima) occurs when the waves are in phase.
The path difference, ? x is an integer multiple of the wavelength (?).
where ?x = n? … (where n = 0 , 1 , 2 , …)
Destructive Interference (Minima) occurs when the waves are out of phase. This happens when
the path difference is a half-integer multiple of the wavelength.
It is given that A is equidistant from both speakers (L
1 A
= L
2 A
). Therefore, the path difference at
A is zero (?x
A
= 0) . This corresponds to the central maximum (n = 0).
As the recorder moves towards B , the distance to L
2
increases more than the distance to L
1
, so
the path difference (?x) steadily increases.
The signal undergoes 10 cycles of minima and maxima. One complete cycle means moving
from one maximum to the next consecutive maximum, which corresponds to an increase in path
difference by exactly one wavelength (?).
Since it goes through 10 full cycles starting from the central maximum, the path difference at
point B must be exactly 10 wavelengths.
?x
B
= 10?
The path length from L
1
to B is,
The path length from L2 to B is,
So, the path difference is,
Read MoreFAQs on Waves: JEE Main Previous Year Questions (2021-2026)
| 1. What are the basic characteristics of waves? |  |
Ans. Waves are disturbances that transfer energy from one point to another without the permanent displacement of the medium. The basic characteristics of waves include wavelength, frequency, amplitude, speed, and period. Wavelength is the distance between successive crests or troughs, frequency refers to the number of waves passing a point per unit time, amplitude is the maximum displacement from the rest position, speed is how fast the wave travels through the medium, and period is the time taken for one complete cycle of the wave.
| 2. What is the difference between transverse and longitudinal waves? | |
Ans. Transverse waves are those in which the particle displacement is perpendicular to the direction of wave propagation, such as in water waves or electromagnetic waves. Longitudinal waves, on the other hand, have particle displacement parallel to the direction of wave propagation, which is characteristic of sound waves. In longitudinal waves, areas of compression and rarefaction are formed as the wave travels through the medium.
| 3. How do sound waves travel through different mediums? | |
Ans. Sound waves can travel through solids, liquids, and gases, but they do so at different speeds due to the density and elasticity of the medium. Generally, sound travels fastest in solids, as the particles are closely packed, allowing quicker transfer of vibrations. In liquids, sound travels slower than in solids but faster than in gases, where the particles are farther apart, resulting in slower transmission of sound. The speed of sound in air at room temperature is approximately 343 m/s.
| 4. What is the principle of superposition in wave interference? | |
Ans. The principle of superposition states that when two or more waves overlap in space, the resultant displacement at any point is the sum of the displacements of the individual waves. This can lead to constructive interference, where waves reinforce each other, resulting in a larger amplitude, or destructive interference, where waves cancel each other out, leading to a smaller amplitude or complete cancellation.
| 5. What are standing waves and how are they formed? | |
Ans. Standing waves are a specific type of wave pattern that appears to be stationary, formed by the interference of two waves of the same frequency and amplitude travelling in opposite directions. This results in fixed nodes (points of no displacement) and antinodes (points of maximum displacement) along the medium. Standing waves are commonly observed in musical instruments, such as strings on a guitar or air columns in wind instruments, where they produce specific frequencies or harmonics.
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