A simple pendulum has a small mass called pendulum bob, which is suspended from a light wire or string.
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 lets start with our Simple Pendulum problems for class 9.
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.
Problem Solution – simple pendulum
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 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 its named as seconds’ pendulum.
A few more problems
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 reason for your answer.
Ratio of their time periods would be 1:1.
Time period of 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
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 ?
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 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.
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