Physics chapters
Oscillations
14 JEE Main questions · Physics
Oscillating systems
- 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)

- (2)

- (3)

- (4)

- (1)
- A simple pendulum of string length 30 cm performs 20 oscillations in 10s. 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)120
- (2)0.75
- (3)7.5
- (4)15
- A cylindrical block of mass and area of cross section 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)
- (3)
- (4)
- A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is , then the force constant of smaller piece is N/m
- (1)15
- (2)20
- (3)60
- (4)45
- 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:

- (1)30
- (2)27
- (3)20
- (4)5
- A uniform disc of radius and mass is free to oscillate about the axis A as shown in the figure. For small oscillations the time period is . ( is acceleration due to gravity)

- (1)
- (2)
- (3)
- (4)
- 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 )
- (1) and
- (2) and 8
- (3) and
- (4) and
- The frequency of oscillation of a mass suspended by spring is . If the length of the spring is cut to half, the same mass oscillates with frequency . The value of is .
- (1)1
- (2)2
- (3)
- (4)
We corrected a printing error in the original paper
Source printed the frequencies with a Latin (, ) for .
SHM basics
- The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is Hz.
- (1)14
- (2)88
- (3)28
- (4)176
- The displacement of a particle, executing simple harmonic motion with time period , is expressed as , where is the amplitude. The maximum value of potential energy of this oscillator is found at . The value of is .
- The equation of motion of a particle is given by cm. The particle will come to rest at time and it will have zero acceleration at time . The and respectively are .
- (1) s, s
- (2) s, s
- (3) s, s
- (4) s, s
- The velocity of a particle executing simple harmonic motion along x-axis is described as , where represents displacement. If the time period of motion is s, the value of is .
We corrected a printing error in the original paper
Source printed the period as “ s, the value of ” ( already denotes the displacement).
- Match List-I with List-II.Choose the correct answer from the options given below.
List-I List-II A. I. Periodic with time period but not simple harmonic motion (SHM) B. II. Periodic with time period but Not SHM C. III. periodic with time period and SHM D. IV. Non-periodic - (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
- A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec. the time required by it to move from to is sec.
- (1)1/4
- (2)5/4
- (3)5/8
- (4)3/8
