Problem Statement
64) The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum?
Solution
f = 1/T = ½ Hz = 0.5 Hz
The time period is 2 sec => That means it’s a seconds’ pendulum
65) A seconds’ pendulum is taken to a place where the acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?
Solution
The time period is inversely proportional to the square root of g
So at the new place, the time period will become double its normal time period.
Here the pendulum is seconds’ pendulum. So normally its time period is 2 secs
So at the new place, it would be double i.e. 4 secs.
66) Find the length of a seconds’ pendulum at a place where g = 10 m/s^2(Take ∏= 3·14).
Solution
So, L = (2^2x 10)/(4 x 9.86) = 1.014 m
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Originally published at https://physicsteacher.in on July 7, 2019.
