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Motion and Time - Measurement of Time and Simple Pendulum

Grade 7CBSE

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

🔑Concepts

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Periodic Motion: A motion that repeats itself at regular intervals of time is called periodic motion. The most common example is the motion of a simple pendulum.

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Simple Pendulum: It consists of a small metallic ball or a piece of stone, called the bob, suspended from a rigid stand by a light, inextensible string.

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Oscillatory Motion: The back-and-forth motion of a simple pendulum is called oscillatory motion. One oscillation is completed when the bob moves from its mean position OO to extreme position AA, then to BB, and back to OO.

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

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Length of Pendulum: The distance from the point of suspension to the center of the bob is known as the length of the pendulum (LL). The time period depends on this length.

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Units of Time: The basic unit of time is the second (ss). Larger units include minutes (minmin) and hours (hh).

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Quartz Clocks: Modern clocks and watches use electronic circuits with one or more crystals of quartz to measure time much more accurately than pendulums.

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Ancient Time-measuring Devices: Before pendulum clocks, people used sundials, water clocks, and sand clocks (hourglasses) to track time based on periodic natural events.

📐Formulae

Time Period(T)=Total Time TakenNumber of Oscillations\text{Time Period} (T) = \frac{\text{Total Time Taken}}{\text{Number of Oscillations}}

Speed=Total DistanceTotal Time Taken\text{Speed} = \frac{\text{Total Distance}}{\text{Total Time Taken}}

Time Taken=DistanceSpeed\text{Time Taken} = \frac{\text{Distance}}{\text{Speed}}

1 hour=3600 seconds1 \text{ hour} = 3600 \text{ seconds}

💡Examples

Problem 1:

A simple pendulum takes 32 s32 \text{ s} to complete 2020 oscillations. What is the time period of the pendulum?

Solution:

T=32 s20=1.6 sT = \frac{32 \text{ s}}{20} = 1.6 \text{ s}

Explanation:

The time period is calculated by dividing the total time taken by the number of oscillations completed. Here, 32 s32 \text{ s} divided by 2020 gives 1.6 s1.6 \text{ s} per oscillation.

Problem 2:

The distance between two stations is 240 km240 \text{ km}. A train takes 44 hours to cover this distance. Calculate the speed of the train in km/h\text{km/h} and m/s\text{m/s}.

Solution:

Speed=240 km4 h=60 km/h\text{Speed} = \frac{240 \text{ km}}{4 \text{ h}} = 60 \text{ km/h} To convert to m/s\text{m/s}: 60×518=30018≈16.67 m/s60 \times \frac{5}{18} = \frac{300}{18} \approx 16.67 \text{ m/s}

Explanation:

First, use the formula Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}. To convert km/h\text{km/h} to m/s\text{m/s}, multiply the value by 518\frac{5}{18} because 1 km/h=1000 m3600 s=518 m/s1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}.

Problem 3:

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

Solution:

Total Time=1 min=60 s\text{Total Time} = 1 \text{ min} = 60 \text{ s} Number of oscillations=Total TimeTime Period=60 s2 s=30\text{Number of oscillations} = \frac{\text{Total Time}}{\text{Time Period}} = \frac{60 \text{ s}}{2 \text{ s}} = 30

Explanation:

First, convert the time into seconds (60 s60 \text{ s}). Since one oscillation takes 2 s2 \text{ s}, divide the total duration by the time period to find the count of oscillations.