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simple pendulum - problems with solution

ICSE, CBSE class 9 physics problems from Simple Pendulum chapter with solution

Last updated on April 19th, 2021 at 05:50 pm

Here we are listing down some numerical physics problems on simple pendulum. Class 9 students of ICSE, CBSE and state boards will find these useful. So let’s start with our Simple Pendulum problems for class 9.

How to solve class 9 physics Problems with Solution from simple pendulum chapter?

Here is a list of problems from this chapter with the solution.
Two simple pendulums A and B have equal length, but their bobs weigh 50 gf and l00 gf respectively. What would be the ratio of their time periods ? Give a reason for your answer.

The ratio of their time periods would be 1:1.
The time period of a simple pendulum is not dependent on the mass of its bob.

Two simple pendulums A and B have length 1·0m and 4·0 m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer.

Time period of A = Ta
and Time period of B = Tb
Ta / Tb =√(1/4) = ½
So Tb = 2 Ta

A will take half of the time taken by B to complete an oscillation.
So obviously pendulum A will make more oscillations in 1 minute.

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


In 60 seconds it makes 40 oscillations
In 1 sec it makes = 40/60 = 2/3 oscillation
So frequency = 2/3 per second = 0.67 Hz
Time period = 1/frequency = 3/2 = 1.5 seconds

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

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

See also  Elevator problems in physics with pseudo force

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?

Time period is inversely proportional to the square root of g
Normal time period = T1 ∞ 1/√g 

Time period at new place = T2 ∞ 1/√(g/4) => T2 ∞ 2/(√g) 

so, T2/T1 = 2

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.

Find the length of a seconds’ pendulum at a place where g = 10 m/s2 (Take  ∏= 3·14).

T = 2 ∏ √(L/g)
so, L = T2 g /(4∏2) …(1)

T = 2 seconds
g = 10 m/s2

So, L = (22 x 10)/(4 x 9.86) = 1.014 m

A pendulum completes 2 oscillations in 5 s. (a) What is its time period? (b) If g = 9·8 m /S^2, find its length.

a)T = time required for one oscillation = 5/2 seconds = 2.5 sec
b)T = 2 ∏ √(L/g)
so, L = T2 g /(4∏2) = (2.52 x 9.8)/[4x 9.86] = 1.55 m

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

T = 2 ∏ √(L/g)
T1= 2 ∏ √(1/g)……(1)
T2= 2 ∏ √(9/g)=3 x 2 ∏ √(1/g)……(2)
T1: T2 =1:3

The time periods of two simple pendulums at a place are in ratio 2:1. What will be the ratio of their length?
T1: T2 =2:1 ……………(1)

As T = 2 ∏ √(L/g)

we can write, T1/T2 = √(L1/L2)

So from equation 1, we can write, √(L1/L2) = 2/1
So, L1/L2 = 4/1
Therefore, the ratio of their lengths is 4:1

It takes 0·2 s for a pendulum bob to move from mean position to one end. What is the time period of the pendulum?

Time period = 4 x the time to move from mean to one end = 0.2 x 4 s = 0.8 s

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 1 second time to move from one extreme to the other. That’s why it’s named as seconds’ pendulum.

See also  Explosion and Conservation of Momentum


What is a simple pendulum? A simple pendulum has a small mass called pendulum bob, which is suspended from a light wire or string.

Note: Some or all of these problems may be listed in our numerical set 1 for class 9. So those question numbers are given here as well for easy tracking.

Go to our listing of numericals


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