/*! 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 14 A simple pendulum has a time per... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 14

A simple pendulum has a time period \(T_{1}\) when on the earth's surface, and \(T_{2}\) when taken to a height \(R\) above the earth's surface, where \(R\) is radius of earth. The value of \(T_{2} / T_{1}\) is (A) 1 (B) \(\sqrt{2}\) (C) 4 (D) 2

Short Answer

Expert verified
The value of \(\frac{T_2}{T_1}\) is 2. The correct answer is (D) 2.

Step by step solution

01

Formula for the time period of a simple pendulum

The formula for 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 gravitational acceleration.
02

Find gravitational acceleration at Earth's surface and height R

On the Earth's surface, the gravitational acceleration is \(g_1 = g\), where g is approximately 9.81 m/s虏. At a height R above the Earth's surface, the gravitational acceleration \(g_2\) can be found using the formula: \[g_2 = \frac{g}{(1 + \frac{R}{R_E})^2}\] where \(R_E\) is the radius of the Earth.
03

Calculate the time periods for the two situations

Using the formula for the time period of a simple pendulum, we have: On Earth's surface: \[T_1 = 2\pi \sqrt{\frac{l}{g}}\] At height R above the Earth's surface: \[T_2 = 2\pi \sqrt{\frac{l}{g_2}}\]
04

Find the ratio \(T_2 / T_1\)

Now we need to find the ratio of the time periods: \[\frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{l}{g_2}}}{2\pi \sqrt{\frac{l}{g}}}\] Simplifying, we get: \[\frac{T_2}{T_1} = \sqrt{\frac{g}{g_2}}\] Now we substitute the formula for \(g_2\) from Step 2: \[\frac{T_2}{T_1} = \sqrt{\frac{g}{\frac{g}{(1 + \frac{R}{R_E})^2}}}\] Further simplification gives: \[\frac{T_2}{T_1} = \sqrt{(1 + \frac{R}{R_E})^2}\] Finally, we find the ratio: \[\frac{T_2}{T_1} = 1 + \frac{R}{R_E}\] Since the height R is equal to the Earth's radius, we have: \[\frac{T_2}{T_1} = 1 + \frac{R_E}{R_E}\] \[\frac{T_2}{T_1} = 1 + 1\] \[\frac{T_2}{T_1} = 2\] The correct answer is (D) 2.

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!

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

The orbital velocity of an artificial satellite in a circular orbit just above earth's surface is \(v_{0} .\) For a satellite orbiting in a circular orbit at an altitude of half of earth's radius is (A) \(\sqrt{\frac{3}{2}} v_{0}\) (B) \(\sqrt{\frac{2}{3}} v_{0}\) (C) \(\frac{3}{2} v_{0}\) (D) \(\frac{2}{3} v_{0}\)

A sphere of density \(\rho\) and radius \(a\) has a concentric cavity of radius \(b\) as shown in the Fig. \(7.13\). Gravitational potential energy as a function of \(r\), where \(r\) is the distance from the centre of the sphere. When \(0

A body of mass \(m\) rises to a height \(h=R / 5\) from the earth's surface where \(R\) is radius of the earth. If \(g\) is acceleration due to gravity at the earth surface, the increase in potential energy is (A) \(m g R\) (B) \((4 / 5) m g R\) (C) \((1 / 6) m g R\) (D) \((6 / 7) m g R\)

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(R\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star is proportional to \(R^{-5 / 2}\), \(T^{2}\) is proportional to (A) \(R^{3}\) (B) \(R^{7 / 2}\) (C) \(R^{3 / 2}\) (D) \(R^{3.75}\)

Four particles, each of mass \(M\) and equidistant from each other, move along a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is (A) \(\sqrt{\frac{G M}{R}}\) (B) \(\sqrt{2 \sqrt{2} \frac{G M}{R}}\) (C) \(\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\) (D) \(\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Classical Mechanics

Read Explanation

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electromagnetism

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.