Problems on Simple Pendulum with solution—

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?


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?


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).


So, L = (2^2x 10)/(4 x 9.86) = 1.014 m

Want a detailed explanation with formula?

Want more such numerical problems with solutions from various chapters? Then please go to the link below.

Get 101+ Numerical Problems with the solution (class 9 / ICSE / high school syllabus)

Originally published at on July 7, 2019.

NIT Grad & IT prof. Writes and teaches HS physics. A motivational speaker and an active blogger with multiple niches. With lots of stories to tell.

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