Frequency of oscillations
1.

In an experiment to measure the period of a simple pendulum, the time for 30 complete oscillations (swings) was found to be 60 s.
(i) Why are 30 oscillations taken ?
(ii) What is the period of the pendulum ?
(iii) What is the frequency of the pendulum ?

Sol:

(i) We measure the time for a large number of oscillations so that a more accurate estimate of the time period may be obtained. The time for one oscillation is too small to be measured accurately.

(ii) Now, Time for 30 oscillations of pendulum = 60 s
so, Time for 1 oscillation of pendulum = (60) / (30)s
= 2 s

Thus, the period of this simple pendulum is 2 seconds.

(iii) In 60 seconds, number of oscillations = 30
So, In 1 second, number of oscillations = 30 / 60
= 0.5

Thus, the frequency of oscillation of this simple pendulum is 0.5 hertz (which is also written as 0.5 Hz).

We could also calculate the frequency of this pendulum from its time–period, by using the relation :
Frequency f = 1 / T
Here, Time–period T = 2 s
So, Frequency f = 1 / 2
= 0.5 Hz
2.

What is the period of a pendulum of length 10.4 metres ? (g = 9.8 m/s2 ; π = 3.14)

Sol:

The period ‘T’ of a pendulum of length ‘L’ is given by the formula :

T = 2 Π√L/g
Here, Π = 3.14,
length of pendulum, L = 10.4 m and
Acceleration due to gravity, g = 9.8 m/s2

Now, putting these values in the above formula, we get :

T = 2 × 3.14 ×√10.4/9.8
= 2 × 3.14 ×√1.06
= 6.28 × 1.029
= 6.465 s

Thus, the time period of this pendulum is 6.28 seconds.

3.

What is the period of a simple pendulum whose length is 3 m ? (g = 9.8m/s2, Π = 3.14)

Sol:

The period T of a simple pendulum is given by the formula

T = 2Π√L/g
Here, Π = 3.14,
L = 3 m
and g = 9.8 m/s2

Putting these values in the above formula, we get :

Time period, T = 2 × 3.14 × √3/9.8
= 2 × 3.14 × 0.55
∴ Time period, T = 3.45 seconds.
4.

The length of a simple pendulum is 76 cm. What will be its time–period if it were taken to the moon where the acceleration due to gravity is one–sixth that on the earth ? (Acceleration due to gravity on the earth is 9.8 m/s 2)

Sol:

Given, acceleration due to gravity on the earth = 9.8 m/s2

So, Acceleration due to gravity on the moon = 9.8 / 6 = 1.63 m/s2 (because it is one–sixth)

Now, the formula for the time–period of a simple pendulum is :

T = 2Π√L/g
Here, Π = 3.14 length of pendulum,
L = 76 cm.
∴ L = 76 / 100 m
= 0.76 m

And, Acceleration due to gravity, g = 1.63 m/s2(calculated above)

Putting these values in the above formula, we get :

T = 2 × 3.14 × √0.76/1.63
= 6.28 × √0.466
= 6.28 × 0.683
= 4.3 s

Thus, the time–period of this simple pendulum on the moon will be 4.3 seconds.

5.

If the length of the pendulum is made four times, the period of pendulum :
(a) is halved.
(b) is doubled.
(c) becomes four times.
(d) becomes one–fourth.
choose the right answer

Sol:

The right answer is (b). That is, if the length of the pendulum is made four times, the period of pendulum is doubled.

This result has been obtained as follows :

The period of a pendulum is directly proportional to the square root of its length.
i.e, T ∝ √L
Now, if the length L is made four times, then it becomes 4L,
so that, T ∝ √4L
T ∝ 2 × √L (Because √4 = 2)
Thus, the period of pendulum becomes 2 times or it gets doubled.

6.

The period of the pendulum depends on its :
(a) Mass.
(b) Length.
(c) Displacement from mean position.
(d) Energy.
Choose the right answer

Sol:

The right answer is (b). That is, the period of the pendulum depends on its length.