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Physical Quantities and Measurement - Measurement of Time

Grade 6ICSE

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

🔑Concepts

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Time is defined as the interval between two instances or events.

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The standard SI unit of time is the second (ss).

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Larger units of time include minutes (minmin), hours (hh), days, months, and years. 11 leap year contains 366366 days.

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A Simple Pendulum consists of a heavy point mass (called the 'bob') suspended by a weightless, inextensible string from a rigid support.

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Oscillation: One complete to-and-fro motion of the bob about its mean position.

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Time Period (TT): The time taken by the pendulum to complete one full oscillation.

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

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Effective Length (ll): The distance from the point of suspension to the center of gravity of the bob.

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Amplitude: The maximum displacement of the bob from its mean position on either side.

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Factors affecting Time Period: The time period of a pendulum depends on its length (ll); it increases with an increase in length. It does not depend on the mass of the bob or the amplitude of oscillation.

📐Formulae

1 min=60 s1 \text{ min} = 60 \text{ s}

1 h=60 min=3600 s1 \text{ h} = 60 \text{ min} = 3600 \text{ s}

1 mean solar day=24 h=86,400 s1 \text{ mean solar day} = 24 \text{ h} = 86,400 \text{ s}

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

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

f=Number of oscillationsTotal time takenf = \frac{\text{Number of oscillations}}{\text{Total time taken}}

💡Examples

Problem 1:

Convert 3 hours 15 minutes3 \text{ hours } 15 \text{ minutes} into seconds (ss).

Solution:

3 h=3×3600 s=10,800 s3 \text{ h} = 3 \times 3600 \text{ s} = 10,800 \text{ s} 15 min=15×60 s=900 s15 \text{ min} = 15 \times 60 \text{ s} = 900 \text{ s} Total time =10,800 s+900 s=11,700 s= 10,800 \text{ s} + 900 \text{ s} = 11,700 \text{ s}

Explanation:

To convert hours to seconds, multiply by 36003600. To convert minutes to seconds, multiply by 6060. Finally, add the two values.

Problem 2:

A simple pendulum completes 5050 oscillations in 2 minutes2 \text{ minutes}. Calculate its Time Period (TT) and Frequency (ff).

Solution:

Total time in seconds =2×60=120 s= 2 \times 60 = 120 \text{ s}. T=Total timeNumber of oscillations=12050=2.4 sT = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{120}{50} = 2.4 \text{ s} f=1T=12.4≈0.416 Hzf = \frac{1}{T} = \frac{1}{2.4} \approx 0.416 \text{ Hz}

Explanation:

First, convert the total time into the SI unit (seconds). The time period is the time per oscillation, and frequency is the reciprocal of the time period.

Problem 3:

If the time period of a simple pendulum is 2 s2 \text{ s}, how many oscillations will it complete in 1 minute1 \text{ minute}?

Solution:

Total time =1 minute=60 s= 1 \text{ minute} = 60 \text{ s}. Time Period (TT) =2 s= 2 \text{ s}. Number of oscillations=Total timeT\text{Number of oscillations} = \frac{\text{Total time}}{T} Number of oscillations=602=30\text{Number of oscillations} = \frac{60}{2} = 30

Explanation:

By rearranging the time period formula, the number of oscillations is found by dividing the total duration by the time taken for one oscillation.