Question

State the numerical value of the frequency of oscillation of a seconds’ pendulum. Does it depend on the amplitude of oscillation ?

February 6, 2024

Answer

As we know that,

f = 1/T

T for a seconds’ pendulum = 2s.

So, substituting the value of T in equation above we get,

or, f = 1/2 = 0.5

Hence, the numerical value of the frequency of oscillation of a seconds’ pendulum is 0.5 s^-1.

No, it does not depend on the amplitude of oscillation.

💡 Some Related Questions

How much time does the bob of a seconds’ pendulum take to move from one extreme of its oscillation to the other extreme?

It takes 0.2s for a pendulum bob to move from mean position to one end. What is the time period of pendulum?

The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the ratio of their lengths ?

A pendulum completes 2 oscillations in 5s. (a) What is its time period? (b) If g = 9.8 ms-2, find its length.

Compare the time periods of two pendulums of length 1m and 9m

Find the length of a seconds’ pendulum at a place where g = 10 ms-2 (Take π = 3.14).

A seconds’ pendulum is taken to a place where acceleration due to gravity falls to one forth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

The time period of a simple pendulum is 2s. What is its frequency? What name is given to such a pendulum?

A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period.

How does the time period (T) of a simple pendulum depend on its length (l) ? Draw a graph showing the variation of T² with l. How will you use this graph to determine the value of g (acceleration due to gravity)?