Home
Class 11
PHYSICS
A simple harmonic motion has an amplitud...

A simple harmonic motion has an amplitude of A and an angular frequency of `omega`. Determine the distance of the point from the equilibrium position where the magnitudes of velocity and acceleration of the motion are equal.

Text Solution

Verified by Experts

The correct Answer is:
`A/sqrt(omega^(2)+1)`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise HOTS NUMERICAL PROBLEMS|22 Videos

  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise ASSERTION-REASON TYPE|3 Videos

  • SIMPLE HARMONIC MOTION

    CHHAYA PUBLICATION|Exercise PROBLEM SET - I|27 Videos

  • ROTATION OF RIGID BODIES

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|10 Videos

  • STATICS

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|3 Videos

Similar Questions

Explore conceptually related problems

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance (2A)/3 from equilibrium position. The new amplitude of the motion is

A moving particle has an acceleration a = -bx, where b is a positive constant and x is the distance of the particle from the equilibrium position. What is the time period of oscillation of the particle?

A particle is in linear simple harmonic motion between two points A and B, 10 cm apart. Take the direction from A to B, as positive direction and give the signs of velocity and acceleration on the particle when it is (i) at the end B, (ii) at 3 cm away from A going towards B.

A particle executes simple harmonic motion along principle axis of a convex lens of focal length f with amplitude ‘a’. The centre point of the motion is at a distance ‘b’ from the lens. Find the amplitude of oscillation of the image.

In case of a simple harmonic motion, if the velocity is plotted along the X-axis and the displacement (from the equilibrium position) is plotted along the Y-axis, the resultant curve happens to be an ellipse with the ratio ("major axis (along X)")/("minor axis (along Y)")=20pi What is the frequency of the simple harmonic motion?

In case of a simple harmonic motion, if the velocity is plotted along the X-axis and the displacement (from the equilibrium position) is plotted along the Y-axis, the resultant curve happens to be an ellipse with the ratio: ("major axis(along X)")/("minor axis(alongY)")=20pi What is the frequency of the simple harmonic motion?

A particle executes SHM with an amplitude of 13 cm and its velocity at a distance of 5 cm from the equilibrium position is 36cm*s^(-1) . Determine the maximum value of the restoring force. (Mass of the particle = 2g).

The maximum acceleration of a particle executing simple harmonic motion is 0.296m*s^(-2) , its time period is 2s and the displacement from the equilibrium position at the beginning of its motion is 0.015m. Determine the equation of motion of the particle.

CHHAYA PUBLICATION-SIMPLE HARMONIC MOTION-PROBLEM SET - II
  1. The equation x=(sqrt3sinpit+cospit)cm represents a simple harmonic mot...

    Text Solution

    |

  2. A particle is subjected to two SHMs in the same direction having equal...

    Text Solution

    |

  3. A simple harmonic motion has an amplitude of A and an angular frequenc...

    Text Solution

    |

  4. Velocities of a particle executing SHM are u and v at distances x(1)an...

    Text Solution

    |

  5. Velocities of a particle executing SHM are 4m*s^(-1)and3m*s^(-1) at di...

    Text Solution

    |

  6. A particle of mass 0.01 kg is executing simple harmonic motion along a...

    Text Solution

    |

  7. Total energy of a particle executing simple harmonic motion is 400 erg...

    Text Solution

    |

  8. The ratio of the masses, time periods and amplitudes of vibration of t...

    Text Solution

    |

  9. A particle of mass m is executing oscillations about the origin of the...

    Text Solution

    |

  10. The mass of the bob of a simple pendulum is 2 g. If the pendulum keeps...

    Text Solution

    |

  11. A seconds pendulum goes slow by 1 min 40 s a day. What should be the c...

    Text Solution

    |

  12. A pendulum clock gives correct time at sea level. By how much time wil...

    Text Solution

    |

  13. The time period of a simple pendulum is 1.96 s. Its time period increa...

    Text Solution

    |

  14. Calculate the percentage of charge of time period of a simple pendulum...

    Text Solution

    |

  15. Two simple harmonic motions are represented by the equations y(1)=0.1s...

    Text Solution

    |

  16. A particle at the end of a spring executes simple harmonic motion with...

    Text Solution

    |

  17. A spring balance has a scale that reads from 0 to 50 kg. The length of...

    Text Solution

    |

  18. If the angular frequency of a particle executing SHM is omega and its ...

    Text Solution

    |

  19. When a particle executing SHM reaches half of its amplitude, find rati...

    Text Solution

    |

  20. The time period and the amplitude of vibration of a particle of mass 5...

    Text Solution

    |