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Measurement - Measurement of Time

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Time is defined as the interval between two events. The standard SI unit of time is the second (ss).

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A simple pendulum consists of a small heavy mass, called the bob, suspended from a rigid support by a weightless and inextensible string.

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One complete to-and-fro motion of the pendulum bob about its mean position is called an oscillation.

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The time taken by the pendulum to complete one oscillation is called its Time Period (TT).

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The number of oscillations completed by the pendulum in one second is called its Frequency (ff). The unit of frequency is Hertz (HzHz).

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The effective length of a pendulum (ll) is the distance from the point of suspension to the center of gravity of the bob.

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The time period of a simple pendulum is directly proportional to the square root of its effective length: T∝lT \propto \sqrt{l}.

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The time period of a simple pendulum is independent of the mass or material of the bob and the amplitude of oscillation (provided the amplitude is small).

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The time period of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity (gg): T∝1gT \propto \frac{1}{\sqrt{g}}.

📐Formulae

T=Total time takenNumber of oscillationsT = \frac{\text{Total time taken}}{\text{Number of oscillations}}

f=1Tf = \frac{1}{T}

T=2πlgT = 2\pi \sqrt{\frac{l}{g}}

T1T2=l1l2\frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}}

💡Examples

Problem 1:

A simple pendulum completes 4040 oscillations in 8080 seconds. Calculate its time period and frequency.

Solution:

Given: Number of oscillations n=40n = 40, Total time t=80 st = 80\text{ s}. Time Period T=tn=8040=2 sT = \frac{t}{n} = \frac{80}{40} = 2\text{ s}. Frequency f=1T=12=0.5 Hzf = \frac{1}{T} = \frac{1}{2} = 0.5\text{ Hz}.

Explanation:

The time period is the duration of one oscillation, while frequency is the number of oscillations per unit time.

Problem 2:

What will happen to the time period of a simple pendulum if its effective length is increased by 44 times?

Solution:

Let the initial length be l1l_1 and final length be l2=4l1l_2 = 4l_1. Using the relation T1T2=l1l2\frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}}, we get: T1T2=l14l1=14=12\frac{T_1}{T_2} = \sqrt{\frac{l_1}{4l_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}. Therefore, T2=2T1T_2 = 2T_1.

Explanation:

Since the time period is proportional to the square root of the length (T∝lT \propto \sqrt{l}), increasing the length fourfold results in doubling the time period.

Problem 3:

A 'seconds pendulum' is a pendulum with a time period of exactly 22 seconds. If the acceleration due to gravity is g=9.8 m/s2g = 9.8\text{ m/s}^2, find the approximate length of a seconds pendulum.

Solution:

Using T=2πlgT = 2\pi \sqrt{\frac{l}{g}}, for T=2 sT = 2\text{ s}: 2=2πl9.82 = 2\pi \sqrt{\frac{l}{9.8}} 1=πl9.81 = \pi \sqrt{\frac{l}{9.8}} Squaring both sides: 1=π2l9.81 = \pi^2 \frac{l}{9.8} l=9.8π2≈9.89.87≈0.992 m≈100 cml = \frac{9.8}{\pi^2} \approx \frac{9.8}{9.87} \approx 0.992\text{ m} \approx 100\text{ cm}.

Explanation:

The length of a seconds pendulum on Earth is approximately 11 meter or 100100 cm.