Home
Class 11
PHYSICS
If length of simple pendulum is increase...

If length of simple pendulum is increased by 6% then percentage change in the time-period will be

A

`3%`

B

`9%`

C

`6%`

D

`1//9 %`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the percentage change in the time period of a simple pendulum when its length is increased by 6%, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Formula for Time Period**: The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. 2. **Identify the Relationship between Length and Time Period**: The time period depends on the square root of the length \( L \). We can express the change in time period in terms of the change in length: \[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \] where \( \Delta T \) is the change in time period and \( \Delta L \) is the change in length. 3. **Convert Changes to Percentage**: To find the percentage change in the time period, we can multiply both sides of the equation by 100: \[ \frac{\Delta T}{T} \times 100 = \frac{1}{2} \frac{\Delta L}{L} \times 100 \] Here, \( \frac{\Delta T}{T} \times 100 \) represents the percentage change in the time period, and \( \frac{\Delta L}{L} \times 100 \) represents the percentage change in length. 4. **Substitute the Given Percentage Change in Length**: We are given that the length of the pendulum is increased by 6%. Therefore: \[ \frac{\Delta L}{L} \times 100 = 6\% \] Substituting this value into the equation gives: \[ \frac{\Delta T}{T} \times 100 = \frac{1}{2} \times 6\% \] 5. **Calculate the Percentage Change in Time Period**: Now, we can calculate: \[ \frac{\Delta T}{T} \times 100 = 3\% \] 6. **Final Answer**: Thus, the percentage change in the time period of the pendulum when the length is increased by 6% is: \[ \text{Percentage change in time period} = 3\% \] ### Summary: The percentage change in the time period of the simple pendulum when its length is increased by 6% is 3%.

To solve the problem of finding the percentage change in the time period of a simple pendulum when its length is increased by 6%, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Formula for Time Period**: The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} ...
Promotional Banner

Topper's Solved these Questions

  • SEMICONDUCTORS

    RESONANCE ENGLISH|Exercise Exercise|29 Videos
  • SOUND WAVES

    RESONANCE ENGLISH|Exercise Exercise- 3 PART - I|47 Videos

Similar Questions

Explore conceptually related problems

The length of a simple pendulum is increased by 44%. The percentage increase in its time period will be

If length of a simple pendulum is increased by 69%, then the percentage increase in its time period is

The length of a simple pendulum is decreased by 21% . Find the percentage change in its time period.

If the length of a simple pendulum is increased by 2%, then the time period

If the length of a simple pendulum is doubled then the % change in the time period is :

The length of a simple pendulum executing simple harmonic motion is increased by 21% . The percentage increase in the time period of the pendulum of increased length is

If the length of a simple pendulum is increased to four times the initial length how is the time period affected?

The length of a simple pendulum is increased four times of its initial valuel, its time period with respect to its previous value will

If the length of a simple pendulum is equal to the radius of the earth, its time period will be

A simple pendulum has some time period T . What will be the percentage change in its time period if its amplitude is decreased by 5%

RESONANCE ENGLISH-SIMPLE HARMONIC MOTION-Exercise
  1. For a simple harmonic vibrator frequency n, the frequency of kinetic e...

    Text Solution

    |

  2. A particle is executing SHM with an amplitude 4 cm. the displacment at...

    Text Solution

    |

  3. For a particle executing S.H.M. which of the following statements hold...

    Text Solution

    |

  4. The equation of SHM of a particle is (d^2y)/(dt^2)+ky=0, where k is a ...

    Text Solution

    |

  5. The total energy of the body executing S.H.M. is E. Then the kinetic e...

    Text Solution

    |

  6. A linear harmonic oscillator of force constant 2 xx 10^(6)N//m and amp...

    Text Solution

    |

  7. A particle executing SHM of amplitude 4 cm and T=4 s . The time taken ...

    Text Solution

    |

  8. The potential energy of a particle execuring S.H.M. is 5 J, when its d...

    Text Solution

    |

  9. A body of mass m is suspended from three springs as shown in figure. I...

    Text Solution

    |

  10. One mass m is suspended from a spring. Time period of oscilation is T....

    Text Solution

    |

  11. A spring has a certain mass suspended from it and its period for verti...

    Text Solution

    |

  12. Two objects A and B of equal mass are suspended from two springs const...

    Text Solution

    |

  13. If the period of oscillation of mass M suspended from a spring is one ...

    Text Solution

    |

  14. A simple pendulum suspended from the ceilling of a stationary trolley ...

    Text Solution

    |

  15. If length of simple pendulum is increased by 6% then percentage change...

    Text Solution

    |

  16. A man measures the period of a simple pendulum inside a stationary lif...

    Text Solution

    |

  17. In case of a forced vibration, the resonance wave becomes very sharp w...

    Text Solution

    |

  18. The amplitude of a damped oscillator becomes half in one minutes. The ...

    Text Solution

    |

  19. Statement-1: kinetic energy of SHM at mean position is equal to potent...

    Text Solution

    |

  20. Statement-1 : Frequency of kinetic energy of SHM is double that of fre...

    Text Solution

    |