Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 18

Pendulum A has a bob of mass \(m\) hung from a string of length \(I_{i}\) pendulum \(B\) is identical to \(A\) except its bob has mass \(2 m\). Compare the frequencies of small oscillations of the two pendulums.

Short Answer

Expert verified
Question: Compare the frequencies of small oscillations of two pendulums, A and B, where pendulum A has a bob of mass m and string length L, and pendulum B has a bob of mass 2m and the same string length L as pendulum A. Answer: The frequencies of small oscillations of pendulums A and B are the same.

Step by step solution

01

Recall the formula for the frequency of a simple pendulum

The formula for the frequency of a simple pendulum is given by: $$ f=\frac{1}{2\pi}\sqrt{\frac{g}{L}} $$ where \(f\) is the frequency, \(g\) is the acceleration due to gravity, and \(L\) is the length of the pendulum.
02

Find the frequency of pendulum A

Since pendulum A has a length of \(L_i\) and a bob of mass \(m\), we can substitute these values into the frequency formula to find its frequency: $$ f_A=\frac{1}{2\pi}\sqrt{\frac{g}{L_i}} $$
03

Find the frequency of pendulum B

Pendulum B has the same length as pendulum A, \(L_i\), but with a bob of mass \(2m\). Since the mass of the bob does not affect the frequency of a simple pendulum, the formula for the frequency of pendulum B is the same as for pendulum A. Hence, the frequency of pendulum B is: $$ f_B=\frac{1}{2\pi}\sqrt{\frac{g}{L_i}} $$
04

Compare the frequencies of small oscillations of the two pendulums

We can now compare the frequencies \(f_A\) and \(f_B\): Since both pendulums have the same length \(L_i\) and the mass of the bob does not affect their frequencies, their frequencies will be the same. $$ f_A=f_B $$ In conclusion, the frequencies of small oscillations of pendulums A and B are the same, regardless of the mass of their bobs.

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

A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will increase the frequency of oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b)

A small mass, \(m=50.0 \mathrm{~g}\), is attached to the end of a massless rod that is hanging from the ceiling and is free to swing. The rod has length \(L=1.00 \mathrm{~m} .\) The rod is displaced \(10.0^{\circ}\) from the vertical and released at time \(t=0\). Neglect air resistance. What is the period of the rod's oscillation? Now suppose the entire system is immersed in a fluid with a small damping constant, \(b=0.0100 \mathrm{~kg} / \mathrm{s},\) and the rod is again released from an initial displacement angle of \(10.0^{\circ}\). What is the time for the amplitude of the oscillation to reduce to \(5.00^{\circ}\) ? Assume that the damping is small Also note that since the amplitude of the oscillation is small and all the mass of the pendulum is at the end of the rod, the motion of the mass can be treated as strictly linear, and you can use the substitution \(R \theta(t)=x(t),\) where \(R=1.0 \mathrm{~m}\) is the length of the pendulum rod.

Two springs, each with \(k=125 \mathrm{~N} / \mathrm{m}\), are hung vertically, and \(1.00-\mathrm{kg}\) masses are attached to their ends. One spring is pulled down \(5.00 \mathrm{~cm}\) and released at \(t=0\); the othen is pulled down \(4.00 \mathrm{~cm}\) and released at \(t=0.300 \mathrm{~s}\). Find the phase difference, in degrees, between the oscillations of the two masses and the equations for the vertical displacements of the masses, taking upward to be the positive direction.

A mass of \(0.404 \mathrm{~kg}\) is attached to a spring with a spring constant of \(206.9 \mathrm{~N} / \mathrm{m}\). Its oscillation is damped. with damping constant \(b=14.5 \mathrm{~kg} / \mathrm{s}\). What is the frequency of this damped oscillation?

14.3 A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring that is pulled to the right \(10.0 \mathrm{~cm}\) and is released from rest. The period of oscillation for the mass is \(5.60 \mathrm{~s}\). What is the speed of the mass at \(t=2.50 \mathrm{~s} ?\) a) \(-2.61 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) b) \(-3.71 \cdot 10^{-2} \mathrm{~m} / \mathrm{s}\) c) \(-3.71 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) d) \(-2.01 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\)
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Fields in Physics

Read Explanation

Radiation

Read Explanation

Mechanics and Materials

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Quantum 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