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 = 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, 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:

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

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 = T^{2} g /(4∏^{2}) = (2.5^{2} 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)

T_{1}= 2 ∏ √(1/g)……(1)

T_{2}= 2 ∏ √(9/g)=3 x 2 ∏ √(1/g)……(2)

T_{1}: T_{2} =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.

**Go to our listing of numericals **

**numerical**

**s**