Home
Class 12
PHYSICS
A simple pendulum of length 1 m is os...

A simple pendulum of length 1 m is oscillating with an angular frequency `10 rad//s`. The suopport of the down with a small angular frequecy of ` 1 rad//s` and an amplitude of `10^(-2)m` . The relative changes in the angular frequency of the pendulum is best given by.

Text Solution

Verified by Experts

Angular frequency of pendulum `omega=sqrt(g/l)`
`:.` relative change in angular frequency
`(Delta omega)/(omega)=1/2(Deltag)/g`[ as length remains constant]
`Deltag=2 A omega_(s)^(2)`[`omega-(2)=` angular frequency of support and A = amplitude]
`(Deltaomega)/(omega)=1/2xx(2 A omega_(s)^(2))/g`
`Delta omega=1/2xx(2xx1^(2)xx10^(-2))/10=10^(-3)` rad/sec
Promotional Banner

Topper's Solved these Questions

  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 14: (WAVES)|15 Videos

  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 15 : (ELECTRIC CHARGES AND FIELDS)|15 Videos

  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 12 (KINETIC THEORY)|15 Videos

  • ATOMS

    DISHA PUBLICATION|Exercise EXERCISE-2: CONCEPT APPLICATOR|30 Videos

  • COMMUNICATION SYSTEM

    DISHA PUBLICATION|Exercise EXERCISE-2 : Concept Applicator|30 Videos

Similar Questions

Explore conceptually related problems

A simple pendulum is oscillating with amplitue A and angular frequency omega At ratio of kinetic energy to potential energy is

A simple pendulum of length 1 m with a bob of mass m swings with an angular amplitude 30^(@) . Then (g= 9.8m//s^(2)

A body is executing the S.H.M. with an angular frequency of 2 rad/sec. Velocity of the body at 20 m displacement, when amplitude of motion is 60 m, is

A ball hung from a vertical spring oscillates in simple harmonic motion with an angular frequency of 2.6 rad/s and an amplitude of 0.075m. What is the maximum acceleration of the ball?

A body is executing simple harmonic motion with an angular frequency s rad/2 . The velocity of the body at 20 mm displacement, when the amplitude of motion is 60 mm , is

A simple pendulum is oscillating with an angular amplitude 60^@ . If mass of bob is 50 gram, then the tension in the string at mean position is (g = 10 ms^(-2))

A simple pendulum of length 40 cm oscillates with an angular amplitude of 0.04 rad. Find a. the time period b. the linear amplitude of the bob, c. The speed of the bob when the strig makes 0.02 rad with the vertical and d. the angular acceleration when the bob is in moemntary rest. Take g=10 ms^-2 .

A simple pendulum is oscillating with amplitude 'A' and angular frequency ' omega ' . At displacement 'x' from mean position, the ratio of kinetic energy to potential energy is