Velocity Displacement Relationship
Trending Questions
Q. A point performs simple harmonic oscillation of period T and the equation of motion is given by x=a sin(ωt+π6). After the elapse of what fraction of the time peirod, the velocity of the point will be equal to half of its maximum velocity?
- T8
- T6
- T3
- T12
Q. The particle is executing SHM on a line 4 cm long. If its velocity at mean position is 12 cm/s, its frequency in Hertz will be :
- 2π3
- 32π
- π3
- 3π
Q. A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum velocity is 30 cm/s, find its velocity when the displacement is 5cm from the mean position.
- 15√3 cm/s
- 20√3 cm/s
- 10√3 cm/s
- 5√3 cm/s
Q. The time period of an object performing SHM is 16 s. It starts its motion from the equilibrium position. After 2 s its velocity is π m/s. What is its displacement amplitude?
- √2 m
- 2√2 m
- 4√2 m
- 8√2 m
Q. A particle executes simple harmonic motion on a straight line. The amplitude of oscillation is 2 cm. At the instant when the displacement of the particle from mean position is 1 cm, the magnitude of acceleration is equal to the magnitude of velocity. The frequency of the simple harmonic motion is
- √37π Hz
- √35π Hz
- √32π Hz
- √34π Hz
Q. The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement ?
Q. A particle is performing SHM with an amplitude A and angular frequency ω. Find the displacement of the particle from the mean position where the speed of the particle becomes half of the maximum speed.
- x=±A2
- x=A√2
- x=±√3A2
- x=−A√2
Q. The particle is executing SHM on a line 4 cm long. If its velocity at mean position is 12 cm/s, its frequency in Hertz will be :
- 2π3
- 32π
- π3
- 3π
Q. A particle executes SHM, with an amplitude of 10 cm. When the particle is at 6 cm from the mean position, the magnitude of its velocity is equal to twice of its acceleration. Find the time period of SHM(in seconds)
- 4.1 s
- 9.42 s
- 1.17 s
- 14.8 s
Q. A particle is executing SHM along a straight line. Its velocities at distances 2 cm and 4 cm from the mean position are 3 cm/s and 2 cm/s respectively. Find the time period of SHM.
- 2π√125 s
- π√125 s
- 2π√65 s
- π√85 s
Q.
The maximum speed and acceleration of a particle executing simple harmonic motion are 10cms−1 and 50cms−2. Find the position(s) of the particle when the speed is 8cm−1.
± 5 cm
± 1.2 cm
± 3.8 cm
± 4.6 cm
Q. A particle describes linear S.H.M. Its velocities are 3ms−1 and 2ms−1 when its displacements are 2 m and 3 m respectively from mean position. Find length of the path?
- 1.2 m
- 4.6 m
- 6.7 m
- 7.2 m
Q. A particle performing SHM has a velocity v0 at mean position. Find the velocity of the particle, when it reaches a point which is at a distance A2 from the mean position.
[A= amplitude of SHM]
[A= amplitude of SHM]
- v04
- √32v0
- 3v02
- v0√2
Q. A particle executes SHM, with an amplitude of 10 cm. When the particle is at 6 cm from the mean position, the magnitude of its velocity is equal to twice of its acceleration. Find the time period of SHM(in seconds)
- 4.1 s
- 9.42 s
- 1.17 s
- 14.8 s
Q. A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum velocity is 30 cm/s, find its velocity when the displacement is 5cm from the mean position.
- 15√3 cm/s
- 20√3 cm/s
- 10√3 cm/s
- 5√3 cm/s
Q. A point performs simple harmonic oscillation of period T and the equation of motion is given by x=a sin(ωt+π6). After the elapse of what fraction of the time peirod, the velocity of the point will be equal to half of its maximum velocity?
- T8
- T6
- T3
- T12
Q. A body executing linear simple harmonic motion has a velocity of 3 cm/s, when its displacement is 4 cm and a velocity of 4 cm/s, when its displacement is 3 cm. What is the amplitude of oscillation?
- 5 cm
- 7.5 cm
- 10 cm
- 12.5 cm