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JEE Exam  >  JEE Notes  >  Physics for JEE Main & Advanced  >  JEE Main Previous Year Questions (2025): Simple Harmonic motion (SHM) and Oscillations

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 Page 1


JEE Main Previous Year Questions 
(2025): Simple Harmonic motion 
(SHM) and Oscillations 
 
Q1: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 
2 times, respectively, than the Earth. The time period of the pendulum remains same 
on earth and the planet. 
Reason ( ?? ) : The mass of the pendulum remains unchanged at Earth and the other 
planet. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both (A) and (R) are true and (R) is the correct explanation of (A) 
B. (A) is false but (R) is true 
C. (A) is true but (R) is false 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: D 
Solution: 
Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times 
and 2 times that of the Earth, respectively, has the same time period as it does on Earth. 
Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other 
planet. 
Explanation: 
The acceleration due to gravity ?? on a planet is given by the formula: 
?? =
????
?? 2
 
where ?? is the gravitational constant, ?? is the mass of the planet, and ?? is the radius of the 
planet. 
For the new planet, the gravitational acceleration ?? '
 is: 
?? '
=
?? ( 4 ?? )
( 2 ?? )
2
=
4 ????
4 ?? 2
=
????
?? 2
= ?? 
Thus, the gravitational acceleration on this new planet is the same as on Earth, ?? . 
The time period ?? of a simple pendulum is determined by: 
Page 2


JEE Main Previous Year Questions 
(2025): Simple Harmonic motion 
(SHM) and Oscillations 
 
Q1: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 
2 times, respectively, than the Earth. The time period of the pendulum remains same 
on earth and the planet. 
Reason ( ?? ) : The mass of the pendulum remains unchanged at Earth and the other 
planet. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both (A) and (R) are true and (R) is the correct explanation of (A) 
B. (A) is false but (R) is true 
C. (A) is true but (R) is false 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: D 
Solution: 
Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times 
and 2 times that of the Earth, respectively, has the same time period as it does on Earth. 
Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other 
planet. 
Explanation: 
The acceleration due to gravity ?? on a planet is given by the formula: 
?? =
????
?? 2
 
where ?? is the gravitational constant, ?? is the mass of the planet, and ?? is the radius of the 
planet. 
For the new planet, the gravitational acceleration ?? '
 is: 
?? '
=
?? ( 4 ?? )
( 2 ?? )
2
=
4 ????
4 ?? 2
=
????
?? 2
= ?? 
Thus, the gravitational acceleration on this new planet is the same as on Earth, ?? . 
The time period ?? of a simple pendulum is determined by: 
?? = 2 ?? v
l
?? 
where l is the length of the pendulum, which indicates that the time period ?? is independent of 
the mass of the pendulum. 
Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly 
explain Assertion (A) because the time period of the pendulum is also independent of the 
pendulum's mass, and the key factor is the unchanged gravitational acceleration. 
Q2: A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is 
pushed down and released to execute simple harmonic oscillations. The time period of 
oscillations is ???? × ????
- ?? ?? , where the value of ?? is(Acceleration due to gravity, ?? =
???? ?? /?? ?? , density of water = ????
?? ???? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 2 
B. 4 
C. 1 
D. 6 
Ans: A 
Solution: 
We can determine the time period of the oscillations by considering that when the cube is 
depressed by a small displacement, the additional buoyant force provided by the displaced 
water acts as a restoring force. Here's a step-by-step explanation: 
Define the given values: 
Side length of the cube, ?? = 10 cm = 0 . 1 m. 
Mass of the cube, ?? = 10 g = 0 . 01 kg. 
Density of water, ?? = 10
3
 kg / m
3
. 
Acceleration due to gravity, ?? = 10 m / s
2
. 
Cross-sectional area of the cube (face area), 
?? = ?? 2
= ( 0 . 1 )
2
= 0 . 01 m
2
. 
When the cube is depressed by a small distance ?? , the additional volume of water displaced is 
? ?? = ???? . 
Thus, the additional buoyant force is: 
?? ?? = ???? ? ?? = ?????? ?? . 
This force acts in the upward (restoring) direction. Notice that the force is proportional to the 
displacement ?? , which is the hallmark of simple harmonic motion (SHM). 
The effective spring constant, ?? , associated with this SHM is: 
Page 3


JEE Main Previous Year Questions 
(2025): Simple Harmonic motion 
(SHM) and Oscillations 
 
Q1: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 
2 times, respectively, than the Earth. The time period of the pendulum remains same 
on earth and the planet. 
Reason ( ?? ) : The mass of the pendulum remains unchanged at Earth and the other 
planet. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both (A) and (R) are true and (R) is the correct explanation of (A) 
B. (A) is false but (R) is true 
C. (A) is true but (R) is false 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: D 
Solution: 
Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times 
and 2 times that of the Earth, respectively, has the same time period as it does on Earth. 
Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other 
planet. 
Explanation: 
The acceleration due to gravity ?? on a planet is given by the formula: 
?? =
????
?? 2
 
where ?? is the gravitational constant, ?? is the mass of the planet, and ?? is the radius of the 
planet. 
For the new planet, the gravitational acceleration ?? '
 is: 
?? '
=
?? ( 4 ?? )
( 2 ?? )
2
=
4 ????
4 ?? 2
=
????
?? 2
= ?? 
Thus, the gravitational acceleration on this new planet is the same as on Earth, ?? . 
The time period ?? of a simple pendulum is determined by: 
?? = 2 ?? v
l
?? 
where l is the length of the pendulum, which indicates that the time period ?? is independent of 
the mass of the pendulum. 
Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly 
explain Assertion (A) because the time period of the pendulum is also independent of the 
pendulum's mass, and the key factor is the unchanged gravitational acceleration. 
Q2: A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is 
pushed down and released to execute simple harmonic oscillations. The time period of 
oscillations is ???? × ????
- ?? ?? , where the value of ?? is(Acceleration due to gravity, ?? =
???? ?? /?? ?? , density of water = ????
?? ???? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 2 
B. 4 
C. 1 
D. 6 
Ans: A 
Solution: 
We can determine the time period of the oscillations by considering that when the cube is 
depressed by a small displacement, the additional buoyant force provided by the displaced 
water acts as a restoring force. Here's a step-by-step explanation: 
Define the given values: 
Side length of the cube, ?? = 10 cm = 0 . 1 m. 
Mass of the cube, ?? = 10 g = 0 . 01 kg. 
Density of water, ?? = 10
3
 kg / m
3
. 
Acceleration due to gravity, ?? = 10 m / s
2
. 
Cross-sectional area of the cube (face area), 
?? = ?? 2
= ( 0 . 1 )
2
= 0 . 01 m
2
. 
When the cube is depressed by a small distance ?? , the additional volume of water displaced is 
? ?? = ???? . 
Thus, the additional buoyant force is: 
?? ?? = ???? ? ?? = ?????? ?? . 
This force acts in the upward (restoring) direction. Notice that the force is proportional to the 
displacement ?? , which is the hallmark of simple harmonic motion (SHM). 
The effective spring constant, ?? , associated with this SHM is: 
?? = ?????? . 
The period of oscillation for SHM is given by: 
?? = 2 ?? v
?? ?? , 
which, upon substituting for ?? , becomes: 
?? = 2 ?? v
?? ?????? . 
Substitute the given values into the expression: 
?? = 2 ?? v
0 . 01
1000 × 10 × 0 . 01
. 
Calculate the denominator: 
1000 × 10 × 0 . 01 = 100 , 
so, 
?? = 2 ?? v
0 . 01
100
= 2 ?? v 0 . 0001 . 
Since, 
v 0 . 0001 = 0 . 01, 
we have: 
?? = 2 ?? × 0 . 01 = 0 . 02 ?? s. 
The problem states that the time period is given by: 
???? × 10
- 2
 s. 
Writing our result in the same form: 
0 . 02 ?? s = 2 ?? × 10
- 2
 s. 
We see that ?? = 2. 
Therefore, the correct answer is: 
Option A: 2. 
Q3: A particle is executing simple harmonic motion with time period 2 s and amplitude 
1 cm . If D and d are the total distance and displacement covered by the particle in 
12.5 s , then 
?? ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 10 
B. 
16
5
 
Page 4


JEE Main Previous Year Questions 
(2025): Simple Harmonic motion 
(SHM) and Oscillations 
 
Q1: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 
2 times, respectively, than the Earth. The time period of the pendulum remains same 
on earth and the planet. 
Reason ( ?? ) : The mass of the pendulum remains unchanged at Earth and the other 
planet. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both (A) and (R) are true and (R) is the correct explanation of (A) 
B. (A) is false but (R) is true 
C. (A) is true but (R) is false 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: D 
Solution: 
Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times 
and 2 times that of the Earth, respectively, has the same time period as it does on Earth. 
Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other 
planet. 
Explanation: 
The acceleration due to gravity ?? on a planet is given by the formula: 
?? =
????
?? 2
 
where ?? is the gravitational constant, ?? is the mass of the planet, and ?? is the radius of the 
planet. 
For the new planet, the gravitational acceleration ?? '
 is: 
?? '
=
?? ( 4 ?? )
( 2 ?? )
2
=
4 ????
4 ?? 2
=
????
?? 2
= ?? 
Thus, the gravitational acceleration on this new planet is the same as on Earth, ?? . 
The time period ?? of a simple pendulum is determined by: 
?? = 2 ?? v
l
?? 
where l is the length of the pendulum, which indicates that the time period ?? is independent of 
the mass of the pendulum. 
Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly 
explain Assertion (A) because the time period of the pendulum is also independent of the 
pendulum's mass, and the key factor is the unchanged gravitational acceleration. 
Q2: A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is 
pushed down and released to execute simple harmonic oscillations. The time period of 
oscillations is ???? × ????
- ?? ?? , where the value of ?? is(Acceleration due to gravity, ?? =
???? ?? /?? ?? , density of water = ????
?? ???? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 2 
B. 4 
C. 1 
D. 6 
Ans: A 
Solution: 
We can determine the time period of the oscillations by considering that when the cube is 
depressed by a small displacement, the additional buoyant force provided by the displaced 
water acts as a restoring force. Here's a step-by-step explanation: 
Define the given values: 
Side length of the cube, ?? = 10 cm = 0 . 1 m. 
Mass of the cube, ?? = 10 g = 0 . 01 kg. 
Density of water, ?? = 10
3
 kg / m
3
. 
Acceleration due to gravity, ?? = 10 m / s
2
. 
Cross-sectional area of the cube (face area), 
?? = ?? 2
= ( 0 . 1 )
2
= 0 . 01 m
2
. 
When the cube is depressed by a small distance ?? , the additional volume of water displaced is 
? ?? = ???? . 
Thus, the additional buoyant force is: 
?? ?? = ???? ? ?? = ?????? ?? . 
This force acts in the upward (restoring) direction. Notice that the force is proportional to the 
displacement ?? , which is the hallmark of simple harmonic motion (SHM). 
The effective spring constant, ?? , associated with this SHM is: 
?? = ?????? . 
The period of oscillation for SHM is given by: 
?? = 2 ?? v
?? ?? , 
which, upon substituting for ?? , becomes: 
?? = 2 ?? v
?? ?????? . 
Substitute the given values into the expression: 
?? = 2 ?? v
0 . 01
1000 × 10 × 0 . 01
. 
Calculate the denominator: 
1000 × 10 × 0 . 01 = 100 , 
so, 
?? = 2 ?? v
0 . 01
100
= 2 ?? v 0 . 0001 . 
Since, 
v 0 . 0001 = 0 . 01, 
we have: 
?? = 2 ?? × 0 . 01 = 0 . 02 ?? s. 
The problem states that the time period is given by: 
???? × 10
- 2
 s. 
Writing our result in the same form: 
0 . 02 ?? s = 2 ?? × 10
- 2
 s. 
We see that ?? = 2. 
Therefore, the correct answer is: 
Option A: 2. 
Q3: A particle is executing simple harmonic motion with time period 2 s and amplitude 
1 cm . If D and d are the total distance and displacement covered by the particle in 
12.5 s , then 
?? ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 10 
B. 
16
5
 
C. 25 
D. 
15
4
 
Ans: C 
Solution: 
?? ( ?? ) = c os ? ( ???? ) 
The particle executes simple harmonic motion (SHM) with amplitude ?? = 1 cm and period ?? =
2 s. In one complete cycle ( 2 s ), the motion is as follows: 
It starts at ?? = 1 cm. 
It moves to ?? = - 1 cm (covering a distance of 2 cm ). 
It returns to ?? = 1 cm (covering another 2 cm ). 
Thus, the total distance traveled in one complete cycle is: 
?? cycle 
= 2 + 2 = 4 cm. 
In 12.5 s , the number of cycles is: 
12 . 5
2
= 6 . 25 cycles. 
For the 6 complete cycles ( 12 s ), the distance traveled is: 
?? complete 
= 6 × 4 = 24 cm. 
For the remaining 0.5 s , determine the displacement. At ?? = 12 s, the particle is at: 
?? ( 12 ) = c os ? ( 12 ?? ) = 1 cm. 
After an extra 0.5 s (i.e., at ?? = 12 . 5 s ), the position becomes: 
?? ( 12 . 5 ) = c os ? ( 12 . 5 ?? ) = c os ? ( 12 ?? +
?? 2
) = c os ?
?? 2
= 0 cm. 
The distance covered in this interval is: 
?? extra 
= | 1 - 0 | = 1 cm. 
So, the total distance traveled is: 
?? = ?? complete 
? + ?? extra 
= 24 + 1 = 25 cm. 
The net displacement, which is the difference between the final and initial positions, is 
calculated as follows: 
Initial position at ?? = 0 : 
?? ( 0 ) = c os ? 0 = 1 cm, 
Final position at ?? = 12 . 5 s : 
?? ( 12 . 5 ) = 0 cm. 
Thus, the displacement is: 
?? = | 0 - 1 | = 1 cm. 
The ratio of the total distance traveled to the displacement is: 
?? ?? =
25
1
= 25. 
Therefore, the correct answer is 25 . 
Q4: A particle oscillates along the ?? -axis according to the law, ?? ( ?? ) = ?? ?? ?? ???? ?? ? (
?? ?? ) 
where ?? ?? = ?? ?? . The kinetic energy ( K ) of the particle as a function of ?? is correctly 
represented by the graph 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
Page 5


JEE Main Previous Year Questions 
(2025): Simple Harmonic motion 
(SHM) and Oscillations 
 
Q1: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 
2 times, respectively, than the Earth. The time period of the pendulum remains same 
on earth and the planet. 
Reason ( ?? ) : The mass of the pendulum remains unchanged at Earth and the other 
planet. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both (A) and (R) are true and (R) is the correct explanation of (A) 
B. (A) is false but (R) is true 
C. (A) is true but (R) is false 
D. Both (A) and (R) are true but (R) is NOT the correct explanation of (A) 
Ans: D 
Solution: 
Assertion (A): A simple pendulum transported to a planet where the mass and radius are 4 times 
and 2 times that of the Earth, respectively, has the same time period as it does on Earth. 
Reason (R): The mass of the pendulum remains unchanged whether on Earth or the other 
planet. 
Explanation: 
The acceleration due to gravity ?? on a planet is given by the formula: 
?? =
????
?? 2
 
where ?? is the gravitational constant, ?? is the mass of the planet, and ?? is the radius of the 
planet. 
For the new planet, the gravitational acceleration ?? '
 is: 
?? '
=
?? ( 4 ?? )
( 2 ?? )
2
=
4 ????
4 ?? 2
=
????
?? 2
= ?? 
Thus, the gravitational acceleration on this new planet is the same as on Earth, ?? . 
The time period ?? of a simple pendulum is determined by: 
?? = 2 ?? v
l
?? 
where l is the length of the pendulum, which indicates that the time period ?? is independent of 
the mass of the pendulum. 
Therefore, while Assertion (A) is true and Reason (R) is true, Reason (R) does not correctly 
explain Assertion (A) because the time period of the pendulum is also independent of the 
pendulum's mass, and the key factor is the unchanged gravitational acceleration. 
Q2: A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is 
pushed down and released to execute simple harmonic oscillations. The time period of 
oscillations is ???? × ????
- ?? ?? , where the value of ?? is(Acceleration due to gravity, ?? =
???? ?? /?? ?? , density of water = ????
?? ???? /?? ?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 2 
B. 4 
C. 1 
D. 6 
Ans: A 
Solution: 
We can determine the time period of the oscillations by considering that when the cube is 
depressed by a small displacement, the additional buoyant force provided by the displaced 
water acts as a restoring force. Here's a step-by-step explanation: 
Define the given values: 
Side length of the cube, ?? = 10 cm = 0 . 1 m. 
Mass of the cube, ?? = 10 g = 0 . 01 kg. 
Density of water, ?? = 10
3
 kg / m
3
. 
Acceleration due to gravity, ?? = 10 m / s
2
. 
Cross-sectional area of the cube (face area), 
?? = ?? 2
= ( 0 . 1 )
2
= 0 . 01 m
2
. 
When the cube is depressed by a small distance ?? , the additional volume of water displaced is 
? ?? = ???? . 
Thus, the additional buoyant force is: 
?? ?? = ???? ? ?? = ?????? ?? . 
This force acts in the upward (restoring) direction. Notice that the force is proportional to the 
displacement ?? , which is the hallmark of simple harmonic motion (SHM). 
The effective spring constant, ?? , associated with this SHM is: 
?? = ?????? . 
The period of oscillation for SHM is given by: 
?? = 2 ?? v
?? ?? , 
which, upon substituting for ?? , becomes: 
?? = 2 ?? v
?? ?????? . 
Substitute the given values into the expression: 
?? = 2 ?? v
0 . 01
1000 × 10 × 0 . 01
. 
Calculate the denominator: 
1000 × 10 × 0 . 01 = 100 , 
so, 
?? = 2 ?? v
0 . 01
100
= 2 ?? v 0 . 0001 . 
Since, 
v 0 . 0001 = 0 . 01, 
we have: 
?? = 2 ?? × 0 . 01 = 0 . 02 ?? s. 
The problem states that the time period is given by: 
???? × 10
- 2
 s. 
Writing our result in the same form: 
0 . 02 ?? s = 2 ?? × 10
- 2
 s. 
We see that ?? = 2. 
Therefore, the correct answer is: 
Option A: 2. 
Q3: A particle is executing simple harmonic motion with time period 2 s and amplitude 
1 cm . If D and d are the total distance and displacement covered by the particle in 
12.5 s , then 
?? ?? is 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 10 
B. 
16
5
 
C. 25 
D. 
15
4
 
Ans: C 
Solution: 
?? ( ?? ) = c os ? ( ???? ) 
The particle executes simple harmonic motion (SHM) with amplitude ?? = 1 cm and period ?? =
2 s. In one complete cycle ( 2 s ), the motion is as follows: 
It starts at ?? = 1 cm. 
It moves to ?? = - 1 cm (covering a distance of 2 cm ). 
It returns to ?? = 1 cm (covering another 2 cm ). 
Thus, the total distance traveled in one complete cycle is: 
?? cycle 
= 2 + 2 = 4 cm. 
In 12.5 s , the number of cycles is: 
12 . 5
2
= 6 . 25 cycles. 
For the 6 complete cycles ( 12 s ), the distance traveled is: 
?? complete 
= 6 × 4 = 24 cm. 
For the remaining 0.5 s , determine the displacement. At ?? = 12 s, the particle is at: 
?? ( 12 ) = c os ? ( 12 ?? ) = 1 cm. 
After an extra 0.5 s (i.e., at ?? = 12 . 5 s ), the position becomes: 
?? ( 12 . 5 ) = c os ? ( 12 . 5 ?? ) = c os ? ( 12 ?? +
?? 2
) = c os ?
?? 2
= 0 cm. 
The distance covered in this interval is: 
?? extra 
= | 1 - 0 | = 1 cm. 
So, the total distance traveled is: 
?? = ?? complete 
? + ?? extra 
= 24 + 1 = 25 cm. 
The net displacement, which is the difference between the final and initial positions, is 
calculated as follows: 
Initial position at ?? = 0 : 
?? ( 0 ) = c os ? 0 = 1 cm, 
Final position at ?? = 12 . 5 s : 
?? ( 12 . 5 ) = 0 cm. 
Thus, the displacement is: 
?? = | 0 - 1 | = 1 cm. 
The ratio of the total distance traveled to the displacement is: 
?? ?? =
25
1
= 25. 
Therefore, the correct answer is 25 . 
Q4: A particle oscillates along the ?? -axis according to the law, ?? ( ?? ) = ?? ?? ?? ???? ?? ? (
?? ?? ) 
where ?? ?? = ?? ?? . The kinetic energy ( K ) of the particle as a function of ?? is correctly 
represented by the graph 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A.  
B.  
C.  
D.  
Ans: D 
Solution: 
?? = ?? 0
s i n
2
?
?? 2
= ?? 0
(
1 - c os ? ?? 2
)
?? -
?? 0
2
=
- c o s ? ?? 2
 where ?? 0
= 1
?? -
1
2
=
- c os ? ?? 2
 
Particle is oscillating between x = 0 to x = 1 
Q5: Given below are two statements. One is labelled as Assertion (A) and the other is 
labelled as Reason (R). 
Assertion (A) : Knowing initial position ?? ?? and initial momentum ?? ?? is enough to 
determine the position and momentum at any time ?? for a simple harmonic motion 
with a given angular frequency ?? . 
Reason ( ?? ) : The amplitude and phase can be expressed in terms of ?? ?? an ?? ?? . 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 28th January Evening Shift 
Options: 
A. (A) is true but (R) is false
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FAQs on JEE Main Previous Year Questions (2025): Simple Harmonic motion (SHM) and Oscillations - Physics for JEE Main & Advanced

1. What is Simple Harmonic Motion (SHM) and how is it characterized?
Ans. Simple Harmonic Motion (SHM) is a type of periodic motion where an object oscillates around an equilibrium position. It is characterized by two main features: the restoring force acting on the object is directly proportional to the displacement from the equilibrium position and is always directed towards that position. Mathematically, this can be expressed as F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement. The motion is sinusoidal in nature, and key parameters include amplitude, period, and frequency.
2. How do you derive the equation of motion for a mass-spring system in SHM?
Ans. To derive the equation of motion for a mass-spring system in SHM, we start with Hooke's law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position: F = -kx. According to Newton's second law, F = ma, where m is the mass and a is the acceleration. Setting these two equations equal gives us ma = -kx. Rearranging leads to a = - (k/m)x. This indicates that the acceleration is proportional to the negative of the displacement, which is the hallmark of SHM. The general solution to this differential equation is x(t) = A cos(ωt + φ), where A is the amplitude, ω = √(k/m) is the angular frequency, and φ is the phase constant.
3. What are the differences between SHM and other types of oscillatory motion?
Ans. The main differences between SHM and other types of oscillatory motion include the nature of the restoring force and the motion characteristics. In SHM, the restoring force is directly proportional to the displacement and always directed towards the equilibrium position. This results in a sinusoidal motion. In contrast, in other oscillatory motions, such as damped oscillations or forced oscillations, the restoring force may not be linear or may involve external forces. Additionally, SHM has a constant amplitude and frequency, while other types of oscillatory motion may have changing amplitude and frequency over time.
4. What role does energy play in Simple Harmonic Motion?
Ans. In Simple Harmonic Motion, energy is continuously converted between kinetic energy and potential energy. At the maximum displacement (amplitude), the potential energy is at its maximum, while the kinetic energy is zero. As the object moves towards the equilibrium position, potential energy decreases while kinetic energy increases. At the equilibrium position, kinetic energy is at its maximum, and potential energy is zero. The total mechanical energy in an ideal SHM system remains constant, assuming no energy is lost to friction or air resistance.
5. How can the concept of SHM be applied in real-life scenarios?
Ans. The concept of SHM can be applied in various real-life scenarios, including mechanical systems such as pendulums, springs, and oscillating masses. For example, clocks utilize pendulums that exhibit SHM to keep accurate time. Additionally, engineers design systems like shock absorbers in vehicles that rely on the principles of SHM to dampen vibrations. In music, instruments like tuning forks produce sound waves that can be modeled as SHM, showcasing its prevalence across different fields.
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Oct 18, 2025 Last updated
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