Numericals of Ex-1(C) of chapter -1 (measurement and experimentation ) of class -9

 1. A simple pendulum complete 40 oscillations in 1. Find its a) frequency, b) time period?

Ans. Simple pendulum completes 40 oscillations in one minute.

Number of oscillations completed in one second.

Time taken to complete one oscillation.

40 oscillations completed=1 minute

40×1 oscillation completed=60 s. (1 minute=60 s)

1 oscillation completed=60 s / 40

One oscillation completed=6/4 s

One oscillation completed=1.5 s

T=1.5 s

1 minute=40 oscillations

60 s=40 oscillations

60×1s=40 oscillations

1s = 40 oscillations / 60

1s=40/60 oscillation

1s=4/6 oscillation

1s=0.67 oscillation

1×s = 0.67 oscillation

1 = 0.67/ s  oscillation

1=0.67 s-¹.    (1/x = x-¹)

By definition of the frequency. we have,

F = 0.67 s-¹.

F=0.67 Hz.

2. The the time period of a simple pendulum is 2s. What is its frequency? What name is given to such a pendulum?

Ans. Time period of a simple pendulum is 2 second.

F=1/T

Frequency=1/2s

                   =0.5 s-¹.  (1/x = x-¹)

We know that, those pendulum whose time period is 2 second are called as seconds pendulum.

3.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?

Ans.a seconds pendulum is taken to a place where acceleration due to gravity falls to one fourth.

T=2π√(l/g)

time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.

So, at a place where ,g=acceleration due to gravity ,l=length, T=time period.

T=2π√(l/g)

Pendulum taken to another place such that

T=2π√( l/ g/4 )

Now time periods ratio wil be as

T/T = 2π√(l/g) / 2π√( l/ g/4 )

T/T = √1/√4

T/T = 1/2

T=(1/2)  T

2 T = T

T= 2 T

T=2× 2s.    (Time period of a seconds pendulum is 2 second, T=2s)

T=4s.

Time period has increased.

New time period is double of the earlier time period. New time period is 4 second.

T is the new time period,g=1/4

 T is the time period,g=1 at a place.

4. Find the length of a seconds pendulum at a place where g =10ms-²( take π=3.14)

Ans.T=2 s

g=10ms-²,π=3.14, l=?

T=2π√(l/g)

Squaring both sides, we get

T×T=2π√(l/g) ×2π√(l/g)

T² =4π²l/g

T² g / 4π² = l

l=T²g / 4π²

Putting the values of T=2s,g=10ms-² & π=3.14 ,we get as

l= (2s)²(10ms-²) / 4(3.14)²

l=4s²×10ms-² / 4×9.8596

l=10m/9.8596

l=1.01423 m

l=1.0142 m

5. Compare the time periods of two pendulums of length one metre and 9 metre.

Ans.l=1m, l=9m

T=2π√(l/g)

T=2π√(l/g).       .......1

T=2π√(l/g).       .......2

Dividing equation 1 by equation 2.

T/T=2π√(l/g) / 2π√(l/g)

T/T=√l/√l

Putting the values of both l in above equation, we get

T/T=√1/√9

T/T=1/3.

6. A pendulum completes 2 oscillations in 5 seconds.a) what is its time period? b) if g=9.8ms-² find its length.

Ans. Pendulum completes 2 oscillations in 5 seconds. 

2 oscillations completed=5s

2× 1 oscillation completed=5s

1 oscillation completed=5s/2

1 oscillation completed=2.5 s

T=2.5 s

T=2π√(l/g)

Squaring both sides ,we get

T²=4π²l/g

T²g / 4π² =l

l=T²g /4π²

Putting the values of T=2.5s,g=9.8ms-²,π=3.14 and solve it.

l=2.5s×2.5s×9.8ms-² /4×3.14×3.14

l=61.25m /39.4384

l=1.55305m

l=1.55m

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

Ans.T/T=2/1.

T=2π√(l/g)

T/T=2π√(l/g) / 2π√(l/g)

T/T=√l/√l

Squaring both sides we get,

(T/T)²=l/l

Putting the ratios of time periods of two simple pendulum as 2: 1

(2/1)²=l/l

4/1=l/l

l/l=4/1

    =4:1

8. It takes 0.2 seconds for a pendulum Bob to move from mean position to one end. What is the time period of pendulum?

Ans. It takes 0.2 second for a pendulum Bob to move from mean position to one end.

But in one oscillation pendulum has to move 4 times from its mean position to one end. Twice to right side and twice to left side.

T=4× time taken to move from mean position to one

T=4×0.2s

T=0.8s

8.how much time does the bob of a seconds pendulum take to move from one extreme of its oscillation to the other extreme?

Ans. Time period of a seconds pendulum =2 s

In one oscillation pendulum has to move 2 times from one extreme of its oscillation to the other extreme.

1 oscillation=2s

2×from one extreme of its oscillation to the other extreme=2s

From one extreme of its oscillation to the other extreme=2s/2

From one extreme of its oscillation to the other extreme=1s.