/*! 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 49 A uniform rod of length \(l\) is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 49

A uniform rod of length \(l\) is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the rod.

Short Answer

Expert verified
The length of the simple pendulum is \( \frac{2l}{3} \).

Step by step solution

01

Understand the Problem

We need to find the length of a simple pendulum that has the same oscillation period as a uniform rod suspended by one end.
02

Time Period of the Rod

The time period \( T \) of a physical pendulum such as a rod of length \( l \) suspended from one end is given by: \[ T = 2\pi \sqrt{\frac{2l}{3g}} \] where \( g \) is the acceleration due to gravity.
03

Time Period of a Simple Pendulum

The time period \( T \) of a simple pendulum with length \( L \) is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \]
04

Equate the Time Periods

To find the equivalent length \( L \) for the simple pendulum, equate the time periods of both pendulums:\[ 2\pi \sqrt{\frac{2l}{3g}} = 2\pi \sqrt{\frac{L}{g}} \]
05

Solve for the Length of the Simple Pendulum

By equating the expressions, cancel \( 2\pi \) from both sides and solve for \( L \):\[ \sqrt{\frac{2l}{3}} = \sqrt{L} \]\[ \frac{2l}{3} = L \]Thus, the equivalent length \( L \) is \( \frac{2l}{3} \).

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.

Uniform Rod Oscillations
Oscillation refers to the repetitive back and forth motion happening around an equilibrium point. In the context of a uniform rod, this refers to a rod pivoted at one end and swinging like a pendulum.
Uniform rods are considered in mechanics to demonstrate oscillatory behavior. The oscillation period depends on the length of the rod and the acceleration due to gravity.
For small oscillations, the rod behaves like a physical pendulum. This type of system helps us understand complex pendulum movements in simple terms and find analogies with simple pendulums.
Time Period Equations
The time period of any pendulum is crucial as it tells us how long it takes for one complete cycle of the swing.
For a physical pendulum like a rod, the equation to determine the time period is \[ T = 2\pi \sqrt{\frac{2l}{3g}} \]Here, \( l \) is the length of the rod and \( g \) is the acceleration due to gravity. This equation considers the distribution of mass along the rod.
In simple pendulums, the period is shorter due to uniform mass distribution being modeled as a point mass at the end of a massless string. The time period for a simple pendulum is represented by \[ T = 2\pi \sqrt{\frac{L}{g}} \]where \( L \) is the pendulum length.
Physical Pendulum
A physical pendulum is a more complex form of the pendulum concept where the mass is distributed rather than concentrated at a point.
In essence, it is any body that swings back and forth under the influence of gravity, not limited to rods. In the case of the uniform rod swinging by one end, it embodies the physical pendulum.
The time period for a physical pendulum is determined by its moment of inertia and pivot point. These factors make the calculation slightly more involved than a simple pendulum but offer a deeper understanding of rotational motion and dynamics.
Equivalent Pendulum Length
Equivalent pendulum length is a concept used to compare a physical pendulum with a simple pendulum. This allows us to find an appropriate simple pendulum that mirrors the oscillation period of a more complex physical pendulum.
To equate a rod with a simple pendulum, set the time periods equations equal. This gives us \[ \frac{2l}{3} = L \]where \( L \) is the length of a simple pendulum with the same time period as the rod.
This calculation provides a direct way to answer how long a simple pendulum should be to mimic a physical pendulum's oscillations, making complex systems simpler to analyze.

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

A simple pendulum of length \(40 \mathrm{~cm}\) is taken inside a deep mine. Assume for the time being that the mine is \(1600 \mathrm{~km}\) deep. Calculate the time period of the pendulum there. Radius of the earth \(=6400 \mathrm{~km}\).

A spring stores 5 J of energy when stretched by \(25 \mathrm{~cm}\). It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?

A rectangular plate of sides \(a\) and \(b\) is suspended from a ceiling by two parallel strings of length \(L\) each (figure 12-E11). The separation between the strings is \(d\). The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Find the time period.

A \(1 \mathrm{~kg}\) block is executing simple harmonic motion of amplitude \(0-1 \mathrm{~m}\) on a smooth horizontal surface under the restoring force of a spring of spring constant \(100 \mathrm{~N} \mathrm{~m}^{-1} . \mathrm{A}\) block of mass \(3 \mathrm{~kg}\) is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

Three simple harmonic motions of equal amplitudes \(A\) and equal time periods in the same direction combine. The phase of the second motion is \(60^{\circ}\) ahead of the first and the phase of the third motion is \(60^{\circ}\) ahead of the second. Find the amplitude of the resultant motion.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Particle Model of Matter

Read Explanation

Physics of Motion

Read Explanation

Oscillations

Read Explanation

Energy Physics

Read Explanation

Magnetism and Electromagnetic Induction

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.