SHM and Oscillations: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Simple Harmonic Motion (Oscillations)  
 
(January 2026) 
 
Q1: As shown in the figure, a spring is kept in a stretched position with some extension 
by holding the masses kg 1 kg and  kg 0.2 kg with a separation more than spring natural 
length and are released. Assuming the horizontal surface to be frictionless, the angular 
frequency (in SI unit) of the system is : 
 
(a) 20 
(b) 5 
(c) 30 
(d) 27 
Ans: (c)  
For a two-body system, the angular frequency (?) is given by: 
 
Where : 
k is the spring constant. 
µ is the reduced mass of the system.  
The reduced mass is calculated using the formula : 
 
Given values of masses are : 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Simple Harmonic Motion (Oscillations)  
 
(January 2026) 
 
Q1: As shown in the figure, a spring is kept in a stretched position with some extension 
by holding the masses kg 1 kg and  kg 0.2 kg with a separation more than spring natural 
length and are released. Assuming the horizontal surface to be frictionless, the angular 
frequency (in SI unit) of the system is : 
 
(a) 20 
(b) 5 
(c) 30 
(d) 27 
Ans: (c)  
For a two-body system, the angular frequency (?) is given by: 
 
Where : 
k is the spring constant. 
µ is the reduced mass of the system.  
The reduced mass is calculated using the formula : 
 
Given values of masses are : 
 
So the angular frequency is, 
 
Therefore, the angular frequency of the system in SI units is 30. Hence, the correct option is (c). 
 
Q2: A cylindrical block of mass M and area of cross section A is floating in a liquid of 
density ? and with its axis vertical. When depressed a little and released the block starts 
oscillating. The period of oscillation is 
(a)  
(b)  
(c)  
(d)  
Ans: (c)  
Sol:  
Initially, the block is floating. The weight of the block is balanced by the buoyant force (F
B
).  
Let h be the height of the submerged part of the block. 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Simple Harmonic Motion (Oscillations)  
 
(January 2026) 
 
Q1: As shown in the figure, a spring is kept in a stretched position with some extension 
by holding the masses kg 1 kg and  kg 0.2 kg with a separation more than spring natural 
length and are released. Assuming the horizontal surface to be frictionless, the angular 
frequency (in SI unit) of the system is : 
 
(a) 20 
(b) 5 
(c) 30 
(d) 27 
Ans: (c)  
For a two-body system, the angular frequency (?) is given by: 
 
Where : 
k is the spring constant. 
µ is the reduced mass of the system.  
The reduced mass is calculated using the formula : 
 
Given values of masses are : 
 
So the angular frequency is, 
 
Therefore, the angular frequency of the system in SI units is 30. Hence, the correct option is (c). 
 
Q2: A cylindrical block of mass M and area of cross section A is floating in a liquid of 
density ? and with its axis vertical. When depressed a little and released the block starts 
oscillating. The period of oscillation is 
(a)  
(b)  
(c)  
(d)  
Ans: (c)  
Sol:  
Initially, the block is floating. The weight of the block is balanced by the buoyant force (F
B
).  
Let h be the height of the submerged part of the block. 
 
 
When the block is depressed by a small distance x downwards, the volume of the displaced 
liquid increases by A · x . This creates an extra buoyant force which acts as the restoring force.  
The additional buoyant force is :  
F
restoring
 = - (Additional Volume) × ? × g  
F
restoring
 = - (Ax) ?g  
Since the force is proportional to displacement x and acts opposite to it (F ? - x), the motion is 
Simple Harmonic Motion (SHM).  
Using Newton's Second Law (F = Ma) :  
Ma = - (? Ag) x  
? a = - (?AgM) x.  
Standard equation for SHM acceleration is a = - ?
2
x  
Comparing the two equations: 
 
The time period T is given by  
Page 4


JEE Main Previous Year Questions (2021-2026): 
Simple Harmonic Motion (Oscillations)  
 
(January 2026) 
 
Q1: As shown in the figure, a spring is kept in a stretched position with some extension 
by holding the masses kg 1 kg and  kg 0.2 kg with a separation more than spring natural 
length and are released. Assuming the horizontal surface to be frictionless, the angular 
frequency (in SI unit) of the system is : 
 
(a) 20 
(b) 5 
(c) 30 
(d) 27 
Ans: (c)  
For a two-body system, the angular frequency (?) is given by: 
 
Where : 
k is the spring constant. 
µ is the reduced mass of the system.  
The reduced mass is calculated using the formula : 
 
Given values of masses are : 
 
So the angular frequency is, 
 
Therefore, the angular frequency of the system in SI units is 30. Hence, the correct option is (c). 
 
Q2: A cylindrical block of mass M and area of cross section A is floating in a liquid of 
density ? and with its axis vertical. When depressed a little and released the block starts 
oscillating. The period of oscillation is 
(a)  
(b)  
(c)  
(d)  
Ans: (c)  
Sol:  
Initially, the block is floating. The weight of the block is balanced by the buoyant force (F
B
).  
Let h be the height of the submerged part of the block. 
 
 
When the block is depressed by a small distance x downwards, the volume of the displaced 
liquid increases by A · x . This creates an extra buoyant force which acts as the restoring force.  
The additional buoyant force is :  
F
restoring
 = - (Additional Volume) × ? × g  
F
restoring
 = - (Ax) ?g  
Since the force is proportional to displacement x and acts opposite to it (F ? - x), the motion is 
Simple Harmonic Motion (SHM).  
Using Newton's Second Law (F = Ma) :  
Ma = - (? Ag) x  
? a = - (?AgM) x.  
Standard equation for SHM acceleration is a = - ?
2
x  
Comparing the two equations: 
 
The time period T is given by  
 
Hence, the correct option is (c). 
 
 
Q3: A spring of force constant 15 N / m is cut into two pieces. If the ratio of their length is 
1 : 3, then the force constant of smaller piece is _____ N / m.  
(a) 20  
(b) 45  
(c) 60  
(d) 15 
Ans: (c)  
Sol: 
The spring constant k of a spring is inversely proportional to the length L of the spring. 
 
This means that the product of the spring constant and the length is a constant for a given 
spring: 
k · L = Constant  
The force constant of original spring is k
o
 = 15 N / m .  
Let the original length be L.  
The spring is cut into two pieces with a length ratio of 1 : 3 . 
 
Hence, the length of smaller spring is L
1
 = x = L/4 and the length of larger spring is L
2
 = 3x = 
3L/4.  
Let k
1
 be the force constant of the smaller piece.  
Using the inverse proportionality relationship: 
 
Therefore, the force constant of smaller piece is 60 N / m.  
Hence, the correct option is (c). 
 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Simple Harmonic Motion (Oscillations)  
 
(January 2026) 
 
Q1: As shown in the figure, a spring is kept in a stretched position with some extension 
by holding the masses kg 1 kg and  kg 0.2 kg with a separation more than spring natural 
length and are released. Assuming the horizontal surface to be frictionless, the angular 
frequency (in SI unit) of the system is : 
 
(a) 20 
(b) 5 
(c) 30 
(d) 27 
Ans: (c)  
For a two-body system, the angular frequency (?) is given by: 
 
Where : 
k is the spring constant. 
µ is the reduced mass of the system.  
The reduced mass is calculated using the formula : 
 
Given values of masses are : 
 
So the angular frequency is, 
 
Therefore, the angular frequency of the system in SI units is 30. Hence, the correct option is (c). 
 
Q2: A cylindrical block of mass M and area of cross section A is floating in a liquid of 
density ? and with its axis vertical. When depressed a little and released the block starts 
oscillating. The period of oscillation is 
(a)  
(b)  
(c)  
(d)  
Ans: (c)  
Sol:  
Initially, the block is floating. The weight of the block is balanced by the buoyant force (F
B
).  
Let h be the height of the submerged part of the block. 
 
 
When the block is depressed by a small distance x downwards, the volume of the displaced 
liquid increases by A · x . This creates an extra buoyant force which acts as the restoring force.  
The additional buoyant force is :  
F
restoring
 = - (Additional Volume) × ? × g  
F
restoring
 = - (Ax) ?g  
Since the force is proportional to displacement x and acts opposite to it (F ? - x), the motion is 
Simple Harmonic Motion (SHM).  
Using Newton's Second Law (F = Ma) :  
Ma = - (? Ag) x  
? a = - (?AgM) x.  
Standard equation for SHM acceleration is a = - ?
2
x  
Comparing the two equations: 
 
The time period T is given by  
 
Hence, the correct option is (c). 
 
 
Q3: A spring of force constant 15 N / m is cut into two pieces. If the ratio of their length is 
1 : 3, then the force constant of smaller piece is _____ N / m.  
(a) 20  
(b) 45  
(c) 60  
(d) 15 
Ans: (c)  
Sol: 
The spring constant k of a spring is inversely proportional to the length L of the spring. 
 
This means that the product of the spring constant and the length is a constant for a given 
spring: 
k · L = Constant  
The force constant of original spring is k
o
 = 15 N / m .  
Let the original length be L.  
The spring is cut into two pieces with a length ratio of 1 : 3 . 
 
Hence, the length of smaller spring is L
1
 = x = L/4 and the length of larger spring is L
2
 = 3x = 
3L/4.  
Let k
1
 be the force constant of the smaller piece.  
Using the inverse proportionality relationship: 
 
Therefore, the force constant of smaller piece is 60 N / m.  
Hence, the correct option is (c). 
 
Q4: A simple pendulum of string length 30 cm performs 20 oscillations in 10 s . The 
length of the string required for the pendulum to perform 40 oscillations in the same time 
duration is _ _ _ _ cm . [Assume that the mass of the pendulum remains same.]  
(a) 0.75  
(b) 7.5  
(c) 15  
(d) 120  
Ans: (b) 
Sol: 
The time period (T) of a simple pendulum is the time taken to complete one full oscillation.  
For a simple pendulum of the length of the string (L) and the acceleration due to gravity (g) the 
formula for the time period is, 
 
So, the time period is directly proportional to the square root of the length 
 
Case - 1  
The initial length of the string is L
1
 = 30 cm  
N
1
 = 20 is the number of oscillations completed in total time t
1
 = 10 s  
So, the initial time period of oscillation is 
 
Case - 2  
N
2
 = 40 is the number of oscillations completed in the same time  
t
2
 = t
1
 = 10 s  
So, the final time period of oscillation is, 
 
We need to find the required length (L
2
) corresponding to the final length of the string, Using the 
proportionality L ? T
2