/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Let the time period of the simpl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 68

Let the time period of the simple pendulum on the earth be 'T'. Then,, \(2 \pi \sqrt{\frac{l_{\mathrm{m}}}{\mathrm{g}_{\mathrm{m}}}}\) where, \(\ell_{\mathrm{m}}, \mathrm{g}_{\mathrm{m}}\) are length of the simple pendulum and acceleration due to gravity on the moon.

Short Answer

Expert verified
Answer: The time period of the simple pendulum on the Moon, T_M, is approximately 2.45 times its time period on Earth, which can be represented as: T_M = T_E * 2.45.

Step by step solution

01

Understand the formula for the time period of a simple pendulum

The formula to calculate the time period of a simple pendulum is given by: \(T = 2 \pi \sqrt{\frac{l}{g}}\) where T is the time period, 'l' is the length of the pendulum, and 'g' is the acceleration due to gravity.
02

Establish the ratio of acceleration due to gravity on the Moon and Earth

The acceleration due to gravity on Earth is approximately \(9.81 m/s^2\) and on the Moon is approximately \(1.63 m/s^2\). Let's represent the acceleration due to gravity on Earth as \(g_E\) and on the Moon as \(g_M\). The ratio of \(g_M\) to \(g_E\) can be written as: \(\frac{g_M}{g_E} = \frac{1.63}{9.81}\)
03

Prepare the formula for the pendulum's time period on the Moon

The formula for the time period of the simple pendulum on the Moon is given by: \(T_M = 2 \pi \sqrt{\frac{l_M}{g_M}}\) Since \(l_M\) is the same as the length on Earth (as we are using the same pendulum), we can directly write the formula for the time period on the Moon as: \(T_M = 2 \pi \sqrt{\frac{l}{g_M}}\)
04

Use the gravity ratio to rewrite the formula for the time period on the Moon

We can rewrite the time period formula on the Moon using the gravity ratio: \(T_M = 2 \pi \sqrt{\frac{l}{g_E \cdot \frac{g_M}{g_E}}}\)
05

Express the time period on the Moon in terms of the time period on Earth

We know that the time period on Earth, \(T_E\), is given by: \(T_E = 2 \pi \sqrt{\frac{l}{g_E}}\) We can rewrite the time period formula on the Moon using the time period on Earth: \(T_M = T_E \cdot \sqrt{\frac{g_E}{g_M}}\)
06

Substitute the acceleration due to gravity ratio and find the time period on the Moon

Now, we can substitute the values of \(\frac{g_M}{g_E}\) and \(T_E\) to find the time period of the pendulum on the Moon, \(T_M\): \(T_M = T_E \cdot \sqrt{\frac{9.81}{1.63}}\) After solving, we get the time period of the simple pendulum on the Moon in terms of its time period on Earth: \(T_M = T_E \cdot 2.45\) So, the time period of the simple pendulum on the Moon is approximately 2.45 times its time period on Earth.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Physics
A simple pendulum consists of a weight suspended from a pivot, allowing it to swing freely back and forth in a regular periodic motion. The motion is due to the gravitational pull acting on the pendulum's mass. Central to understanding this motion is the time period of the pendulum, which is the time it takes for the pendulum to complete one full swing, returning to its starting position.

The formula to calculate this time period is . where 'T' represents the time period, 'l' the length of the pendulum, and 'g' the acceleration due to gravity. The time period is independent of the mass of the pendulum and its amplitude, provided the swings are small.

For students aiming to enhance their comprehension, it is pivotal to conduct experiments with pendulums of varying lengths and observe the impact on the time period. Visualization tools, such as pendulum simulations, can also cement understanding of the principles governing pendulum movement.
Acceleration Due to Gravity
The acceleration due to gravity ('g'), is the rate at which an object accelerates when falling freely towards a celestial body, such as Earth or the Moon. On Earth, this value is approximately , imparting a sense of weight to objects. This acceleration is instrumental in determining the motion of pendulums and other freely falling objects.

Understanding 'g' is crucial in many physics problems. Students are advised to employ this concept intuitively, thinking about how the strength of gravity affects an object's acceleration when dropped, or regarding the resultant forces on a body resting on Earth's surface. Graphical representations of free fall in different gravitational strengths can significantly help in visualizing how gravity influences an object's movement.
Comparing Earth and Moon Gravity
When comparing gravity on Earth and the Moon, we find a significant difference. With the Moon's gravity only about of Earth's, this disparity hugely impacts the time period of a pendulum.

This comparison vividly illustrates the concept of varying gravitational forces across celestial bodies. Students can deepen their understanding by researching how the gravity on different planets affects the weight and motion of objects. Engaging in exercises that involve calculating the impact of different gravitational strengths on the same object, like our pendulum, can solidify this concept. Exploring how astronauts' movements differ on the moon compared to earth due to reduced gravity can also be a fascinating and educational exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fill in the Blanks. \(7200 \mathrm{~s}\) 2 hour \(=2 \times 60\) minutes \(=2 \times 60 \times 60 \mathrm{~s}=7200 \mathrm{~s}\)

Distance travelled in the first \(20 \mathrm{~min}=\) speed \(\times\) time \(=60 \times \frac{20}{60}=20 \mathrm{~km}\)

From the given figure, by pythogorous theorem \(\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{R} \mathrm{Q}^{2}\) \(\mathrm{PR}^{2}=4^{2}+3^{2}\) \(\mathrm{PR}^{2}=16+9\) \(\mathrm{PR}=\sqrt{25}\) \(\mathrm{PR}=5 \mathrm{~m}\) Average speed \(=\frac{\text { total distance }}{\text { total time }}=\frac{4 \mathrm{~m}+3 \mathrm{~m}+5 \mathrm{~m}}{20 \mathrm{~s}+10 \mathrm{~s}+30 \mathrm{~s}}=\frac{12 \mathrm{~m}}{60 \mathrm{~s}}=0.2 \mathrm{~m} \mathrm{~s}^{-1}\)

The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)

The different kinds of motion are: (i) Translatory motion: \(\mathrm{A}\) bus moving on a road, the motion of a rising balloon, the free fall of a stone under gravity, the motion of a cricket ball when it is hit by a batsman are examples of translatory motion. Translatory motion is further classified as rectilinear motion and curvilinear motion. When an object moves along a straight path, its motion is said to be rectilinear motion. The marching of soldiers on a straight road, the motion of a car on a straight road, the motion of carrom board coin are examples of rectilinear motion. When an object moves along curved path, its motion is called curvilinear motion. A bus moving on a fly-over bridge, a car taking a turn, a football kicked from the ground into air all have curvilinear motion. (ii) Rotatory motion: In this type of motion, the object rotates about a fixed axis. The motion of blades of a ceiling fan, the spin motion of a top, the motion of turbine, the motion of the earth around the sun are all circular or rotatory motion. In some cases, the rotatory and translatory motions take place simultaneously. When a bicycle moves, its wheels undergo translatory and rotatory motion. (iii) Oscillatory motion: A boy on a swing moves to-and-fro (back and forth). The motion such as above, where an object moves to-and-fro is called oscillatory motion. Other examples of oscillatory motion are the motion of the pendulum of a clock, the motion of a needle of a sewing machine, the motion of a piston of an engine etc.
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Classical Mechanics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.