# Problems on Simple Pendulum with solution— PhysicsTeacher.in

# 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.*