## 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

Time period is 2 sec => That means it’s a seconds’ pendulum

65) A seconds’ pendulum is taken to a place where 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

Time period is inversely proportional to the square root of g

Normal time period = T_{1} ∞ 1/√g

Time period at new place = T_{2 }∞ 1/√(g/4) => T_{2 }∞ 2/(√g)

so, T_{2}/T_{1} = 2

So at the new place time period will become double of 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

T = 2 ∏ √(L/g)

so, L = T^{2} g /(4∏^{2}) …(1)

T = 2 seconds

g = 10 m/s^{2}

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

**Go back to the 101 problems set (class 9)**

**Problems Set – grade 9**