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Oscillations · Question 3

JEE Main 2026 (Online) 6th April Evening ShiftMCQMedium📈 Topic in 16% papers+4−1
1

A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are Hz and J, respectively. (Take g = 10 m/s2)

  1. 5 50 and 8 10-3
  2. 5 50 and 8
  3. 10 50 and 2 10-3
  4. 5 50 and 16 10-3

Answer

(1) 5 50 and 8 10-3

Solution

Spring constant from the static stretch: k= mgx0= 0.2 102 10-3=1000 N/m

Frequency: f= 12 km= 12 10000.2= 50002

⇒ f= 10 502 = 5 50 Hz

The mass oscillates with amplitude 2 mm about the equilibrium position, so at the lowest point the spring is stretched by xmax=2+2=4 mm

∴ Umax= 12kxmax2= 12 1000 (4 10-3 )2

Umax=8 10-3 J

Answer: (1)

JEE Main 2026 (Online) 6th April Evening Shift

Keep practising — Oscillations

2

Match List-I with List-II. Choose the correct answer from the options given below :

  1. A-III, B-I, C-IV, D-II
  2. A-II, B-I, C-III, D-IV
  3. A-III, B-II, C-IV, D-I
  4. A-II, B-I, C-IV, D-III
JEE Main 2026 (Online) 5th April Evening Shift
3

The velocity of a particle executing simple harmonic motion along x-axis is described as v2=50-x2, where x represents displacement. If the time period of motion is x7 s, the value of x is .

JEE Main 2026 (Online) 4th April Morning Shift
4

A uniform disc of radius R and mass M is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is . (g is acceleration due to gravity) [figure]

  1. 2 5R4g
  2. 2 2R3g
  3. 2 3R2g
  4. 2 3Rg
JEE Main 2026 (Online) 4th April Evening Shift
5

The equation of motion of a particle is given by x=a (50t+ /3) cm. The particle will come to rest at time t1 and it will have zero acceleration at time t2. The t1 and t2 respectively are .

  1. 300 s, 75 s
  2. 75 s, 300 s
  3. 300 s, 25 s
  4. 50 s, 100 s
JEE Main 2026 (Online) 2nd April Evening Shift
6

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A t, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T/2 . The value of is .

JEE Main 2026 (Online) 28th January Morning Shift
7

As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 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 : [figure]

  1. 20
  2. 5
  3. 30
  4. 27
JEE Main 2026 (Online) 28th January Evening Shift
8

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

  1. 2 AMg
  2. 2M Ag
  3. 2 M Ag
  4. AMg
JEE Main 2026 (Online) 24th January Morning Shift
9

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their lengths is 1 : 3, then the force constant of smaller piece is N/m.

  1. 20
  2. 45
  3. 60
  4. 15
JEE Main 2026 (Online) 24th January Morning Shift
10

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.]

  1. 0.75
  2. 7.5
  3. 15
  4. 120
JEE Main 2026 (Online) 23rd January Morning Shift
11

Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 22nd January Evening Shift

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