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.
61)
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.

Solution:
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.

62)
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.

Solution:
Time period of A = T_a
and Time period of B = T_b
T_a / T_b =√(1/4) = ½
So T_b = 2 T_a

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.

63)
A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period.

Solution:

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

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 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:
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.

66)
Find the length of a seconds’ pendulum at a place where g = 10 m/s² (Take  ∏= 3·14).

Solution:
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

67)
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.

Solution:
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

68)
Compare the time periods of two pendulums of length 1 m and 9 m

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

69)
The time periods of two simple pendulums at a place are in ratio 2:1. What will be the ratio of their length?
Solution:
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

70)
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?

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

71)
How much time does the bob of a seconds’ pendulum take to move from one extreme of its oscillation to the other extreme?

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

FAQ

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.

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