Home
Class 12
PHYSICS
The Shortest distance travelled by a par...

The Shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to `sqrt(3)//2` times its amplitude . Determine its time period.

Text Solution

Verified by Experts

The correct Answer is:
A, B

`:' y = Asinomegat = Asin'(2pi)/(T) t`
`:. (sqrt(3))/(2) A = Asin((2pi)/(T) xx 2) = Asin ((4pi)/(T))`
`rArr (4pi)/(T) = (pi)/(3) rArr T = 12 s`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-04 [B]|105 Videos
  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-05 [A]|39 Videos
  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Comprehension Base Questions (5)|3 Videos
  • RACE

    ALLEN|Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|24 Videos
  • TEST PAPER

    ALLEN|Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

The shortest distance travelled by a particle executing S.H.M. from mean position in 4 seconds is eual to (1)/(sqrt(2)) times its amplitude. Find the time period. Hint : t=4 s , x = (1)/(sqrt(2)) , T=? x = A sin omega t , omega = ( 2pi)/( T ) Solve to get T.

A particle executes SHM from extreme position and covers a distance equal to half to its amplitude in 1 s. find out it's Time Period.

The potential energy of a particle executing SHM change from maximum to minimum in 5 s . Then the time period of SHM is:

The time period of a particle executing S.H.M.is 12 s. The shortest distance travelledby it from mean position in 2 second is ( amplitude is a )

A particle executing SHM with time period of 2 s : Find the time taken by it to move from one amplitude to half the amplitude position.

The particle is executing SHM on a line 4 cm long. If its velocity at mean position is 12 m/s , then determine its frequency.

A particle executes linear simple harmonic motion with an amplitude of 3 cm . When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is

A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

The kinetic energy of a particle, executing S.H.M. is 16 J when it is in its mean position. IF the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 Kg, the time period of its oscillations is

ALLEN-SIMPLE HARMONIC MOTION-Exercise-04 [A]
  1. A particle simple harmonic motion completes 1200 oscillations per minu...

    Text Solution

    |

  2. Find the resulting amplitude and phase of the vibrations s=Acosomega...

    Text Solution

    |

  3. A particle is executing SHM given by x = A sin (pit + phi). The initia...

    Text Solution

    |

  4. The Shortest distance travelled by a particle executing SHM from mean ...

    Text Solution

    |

  5. Two particle A and B execute SHM along the same line with the same amp...

    Text Solution

    |

  6. A body executing S.H.M. has its velocity 10cm//s and 7 cm//s when its ...

    Text Solution

    |

  7. A particle executing a linear SHM has velocities of 8 m/s 7 m/s and 4 ...

    Text Solution

    |

  8. A particle is oscillating in a straight line about a centre O, with a ...

    Text Solution

    |

  9. The displacement of a particle varies with time as x = 12 sin omega t ...

    Text Solution

    |

  10. A particle of mass 0.1 kg is executing SHM of amplitude 0.1 m . When t...

    Text Solution

    |

  11. A body of mass 1 kg suspended an ideal spring oscillates up and down. ...

    Text Solution

    |

  12. The potential energy (U) of a body of unit mass moving in a one-di...

    Text Solution

    |

  13. A body of mass 1.0 kg is suspended from a weightless spring having for...

    Text Solution

    |

  14. In the figure shown, the block A of mass m collides with the identical...

    Text Solution

    |

  15. A block of mass 1kg hangs without vibrations at the end of a spring wi...

    Text Solution

    |

  16. A small ring of mass m(1) is connected by a string of length l to a sm...

    Text Solution

    |

  17. Calculate the time period of a uniform square plate of side 'a' if it ...

    Text Solution

    |

  18. Two identical rods each of mass m and length L, are tigidly joined and...

    Text Solution

    |

  19. A half ring of mass m, radius R is hanged at its one end its one end i...

    Text Solution

    |

  20. The two torsion pendula differ only by the addition of cylindrical mas...

    Text Solution

    |