Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 61

A pendulum of length \(L_{1}\) has a period \(T_{1}=0.950 \mathrm{s}\). The length of the pendulum is adjusted to a new value \(L_{2}\) such that $T_{2}=1.00 \mathrm{s} .\( What is the ratio \)L_{2} / L_{1} ?$

Short Answer

Expert verified
Answer: The ratio of the adjusted length to the initial length of the simple pendulum is approximately 1.11.

Step by step solution

01

Recalling the formula for the period of a simple pendulum

The formula for the period \(T\) of a simple pendulum of length \(L\) and small oscillation amplitude is given by: $$ T = 2 \pi \sqrt{\frac{L}{g}} $$ where \(g\) is the acceleration due to gravity (\(\approx 9.81 \mathrm{m/s^2}\)).
02

Applying the formula to the initial length and period

Given the initial period \(T_1 = 0.950 \ \mathrm{s}\) and length \(L_1\), we apply the formula for the period of a simple pendulum: $$ T_1 = 2 \pi \sqrt{\frac{L_1}{g}} $$
03

Applying the formula to the adjusted length and period

Given the final period \(T_2 = 1.00 \ \mathrm{s}\) and the new length \(L_2\), we apply the formula for the period of a simple pendulum: $$ T_2 = 2 \pi \sqrt{\frac{L_2}{g}} $$
04

Writing the ratio of the two periods

Let's divide the equation for \(T_2\) by the equation for \(T_1\), which gives us the ratio of the two periods: $$ \frac{T_2}{T_1} = \frac{2 \pi \sqrt{\frac{L_2}{g}}}{2 \pi \sqrt{\frac{L_1}{g}}} $$
05

Simplifying the expression

Cancelling the common terms and cross-multiplying, we have: $$ \frac{T_2}{T_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}} $$ Now, square both sides of the equation to get rid of the square roots: $$ \left(\frac{T_2}{T_1}\right)^2 = \frac{L_2}{L_1} $$
06

Solving for the ratio \(L_2 / L_1\)

The problem asks for the ratio \(L_2 / L_1\). Plug the given values of \(T_1\) and \(T_2\) into the equation above: $$ \frac{L_2}{L_1} = \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{1.00 \ \mathrm{s}}{0.950 \ \mathrm{s}}\right)^2 $$ Now, calculate the ratio: $$ \frac{L_2}{L_1} \approx 1.11 $$ So the ratio \(L_2 / L_1\) is approximately \(1.11\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Resilin is a rubber-like protein that helps insects to fly more efficiently. The resilin, attached from the wing to the body, is relaxed when the wing is down and is extended when the wing is up. As the wing is brought up, some elastic energy is stored in the resilin. The wing is then brought back down with little muscular energy, since the potential energy in the resilin is converted back into kinetic energy. Resilin has a Young's modulus of $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ (a) If an insect wing has resilin with a relaxed length of \(1.0 \mathrm{cm}\) and with a cross-sectional area of \(1.0 \mathrm{mm}^{2},\) how much force must the wings exert to extend the resilin to \(4.0 \mathrm{cm} ?\) (b) How much energy is stored in the resilin?
A bob of mass \(m\) is suspended from a string of length \(L\) forming a pendulum. The period of this pendulum is \(2.0 \mathrm{s}\) If the pendulum bob is replaced with one of mass \(\frac{1}{3} m\) and the length of the pendulum is increased to \(2 L\), what is the period of oscillation?
A small bird's wings can undergo a maximum displacement amplitude of $5.0 \mathrm{cm}$ (distance from the tip of the wing to the horizontal). If the maximum acceleration of the wings is \(12 \mathrm{m} / \mathrm{s}^{2},\) and we assume the wings are undergoing simple harmonic motion when beating. what is the oscillation frequency of the wing tips?
A horizontal spring with spring constant of \(9.82 \mathrm{N} / \mathrm{m}\) is attached to a block with a mass of \(1.24 \mathrm{kg}\) that sits on a frictionless surface. When the block is 0.345 m from its equilibrium position, it has a speed of \(0.543 \mathrm{m} / \mathrm{s}\) (a) What is the maximum displacement of the block from the equilibrium position? (b) What is the maximum speed of the block? (c) When the block is \(0.200 \mathrm{m}\) from the equilibrium position, what is its speed?
An object of mass \(306 \mathrm{g}\) is attached to the base of a spring, with spring constant \(25 \mathrm{N} / \mathrm{m},\) that is hanging from the ceiling. A pen is attached to the back of the object, so that it can write on a paper placed behind the mass-spring system. Ignore friction. (a) Describe the pattern traced on the paper if the object is held at the point where the spring is relaxed and then released at \(t=0 .\) (b) The experiment is repeated, but now the paper moves to the left at constant speed as the pen writes on it. Sketch the pattern traced on the paper. Imagine that the paper is long enough that it doesn't run out for several oscillations.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Oscillations

Read Explanation

Electricity

Read Explanation

Mechanics and Materials

Read Explanation

Thermodynamics

Read Explanation

Fields in Physics

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.

Sign-up for free