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Chapter 14: Problem 65

Two pendulums of identical length of \(1.000 \mathrm{~m}\) are suspended from the ceiling and begin swinging at the same time. One is at Manila, in the Philippines, where \(g=9.784 \mathrm{~m} / \mathrm{s}^{2}\) and the other is at Oslo, Norway, where \(g=9.819 \mathrm{~m} / \mathrm{s}^{2},\) After how many oscillations of the Manila pendulum will the two pendulums be in phase again? How long will it take for them to be in phase again?

Short Answer

Expert verified
Answer: To find the number of oscillations of the Manila pendulum when the two pendulums will be in phase again and the time it will take for them to be in phase again, follow these steps: 1. Compute the period of each pendulum using the formula \(T = 2\pi\sqrt{\frac{l}{g}}\), where T is the period, l is the length of the pendulum, and g is the gravitational acceleration. 2. Find the ratio of the periods, \(\frac{T_1}{T_2} = \frac{\sqrt{g_2}}{\sqrt{g_1}}\). 3. Calculate the smallest integer value for n and m that satisfy the equation: \(n \frac{\sqrt{g_2}}{\sqrt{g_1}} \approx m\) 4. Calculate the time for the pendulums to be in phase again using the formula: \(\text{Time to be in phase again} = nT_1 = n\cdot 2\pi\sqrt{\frac{l_1}{g_1}}\).

Step by step solution

01

Compute the period of each pendulum

We can use the formula for the period of a simple pendulum: $$ T = 2\pi\sqrt{\frac{l}{g}} $$ Where T is the period, l is the length of the pendulum, and g is the gravitational acceleration. For Manila, we are given length \(l_1 = 1.000 \,\mathrm{m}\) and gravitational acceleration \(g_1 = 9.784\, \mathrm{m}/\mathrm{s^2}\). For Oslo, we are given length \(l_2 = 1.000\, \mathrm{m}\) and gravitational acceleration \(g_2 = 9.819\, \mathrm{m}/\mathrm{s^2}\). Compute the period for each pendulum: $$ T_1 = 2\pi\sqrt{\frac{l_1}{g_1}} $$ $$ T_2 = 2\pi\sqrt{\frac{l_2}{g_2}} $$
02

Find the ratio of the periods

To find when the pendulums will be in phase again, we need to find the ratio of the periods. Since we want to find the number of oscillations of the Manila pendulum, we can find the ratio as follows: $$ \frac{T_1}{T_2} = \frac{2\pi\sqrt{\frac{l_1}{g_1}}}{2\pi\sqrt{\frac{l_2}{g_2}}} $$ Simplify the equation as the length of both pendulums is the same, \(l_1 = l_2\): $$ \frac{T_1}{T_2} = \frac{\sqrt{g_2}}{\sqrt{g_1}} $$
03

Calculate the number of oscillations

Now, it's time to calculate the number of oscillations of the Manila pendulum when the two pendulums are in phase again. To do this, we need to find the smallest integer n satisfying: $$ n\frac{T_1}{T_2} \approx m $$ Where m is an integer. Since we already have the expression for \(\frac{T_1}{T_2}\) from Step 2, we can plug it in: $$ n \frac{\sqrt{g_2}}{\sqrt{g_1}} \approx m $$ Find the smallest integer value for n and m that satisfy the equation.
04

Calculate the time for the pendulums to be in phase again

Now that we have the number of oscillations of the Manila pendulum, it's time to calculate the time it takes for both pendulums to be in phase again. We can use the period of the Manila pendulum and multiply it by the number of oscillations: $$ \text{Time to be in phase again} = nT_1 = n\cdot 2\pi\sqrt{\frac{l_1}{g_1}} $$ Calculate the time using the value of n obtained in Step 3.

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Key Concepts

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

Simple Pendulum Period
The period of a simple pendulum is a fundamental concept in physics, describing the time it takes for the pendulum to complete one full oscillation, or swing back and forth once. The formula \[ T = 2\pi\sqrt{\frac{l}{g}} \]is vital to understanding pendulum motion, where \( T \) is the period, \( l \) is the length of the pendulum, and \( g \) is the gravitational acceleration. For a pendulum swinging in a consistent and uninterrupted motion, this period remains constant depending on the pendulum's length and the gravitational acceleration at its location.Understanding the period is crucial because it helps in predicting the behavior of pendulums. For example, clockmakers use this knowledge to design pendulum clocks with accurate timekeeping. In our exercise, calculating the period allowed us to compare the oscillations of pendulums in different gravitational fields.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object accelerates due to the gravitational force of a much larger body, like Earth. It is denoted by \( g \) and is usually measured in meters per second squared (\( \mathrm{m}/\mathrm{s^2} \)). On the surface of the Earth, \( g \) varies slightly depending on altitude and geographical location but is approximately 9.8 \( \mathrm{m}/\mathrm{s^2} \).
In our exercise, the difference in gravitational acceleration between Manila and Oslo slightly changes the pendulums' periods, affecting when they will be in phase again. It's intriguing to see how a small difference in \( g \) values leads to tangible differences in pendulum synchronicity. This phenomenon has real-world implications, such as influencing the accuracy of sensitive equipment and scientific measurements.
Oscillation Phase Relation
Oscillation phase relation refers to the relative position in the oscillation cycle of two or more oscillating objects. When the objects reach the same position at the same time during their cycles, they are said to be 'in phase.' Conversely, when they reach opposite positions, they are 'out of phase.'
In this context, understanding the oscillation phase relation helped us determine after how many oscillations the Manila pendulum and the Oslo pendulum would realign. By equating the periods and solving for a common multiple, we found the point at which both pendulums would return to their starting position simultaneously, indicating they're in phase again. This principle of phase relation in oscillations is crucial not just for pendulums but also in various fields including electronics, acoustics, and even quantum physics.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is proportional to the displacement and acts in the opposite direction. It's characterized by oscillations around an equilibrium position, with the motion being symmetrical and occurring at regular intervals. Pendulums exhibit SHM when their oscillations are small, which is why the formulas we've been using apply.
The pendulum exercise embodies the properties of SHM, with gravity providing the restoring force. It's noteworthy that SHM can be used to model various physical systems beyond pendulums, like springs, certain molecular vibrations, and even the oscillation of celestial bodies in some cases. The concept of SHM is fundamental for understanding a wide range of physical phenomena and engineering systems.

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Most popular questions from this chapter

Cars have shock absorbers to damp the oscillations that would otherwise occur when the springs that attach the wheels to the car's frame are compressed or stretched. Ideally, the shock absorbers provide critical damping. If the shock absorbers fail, they provide less damping, resulting in an underdamped motion. You can perform a simple test of your shock absorbers by pushing down on one corner of your car and then quickly releasing it If this results in an up-and- down oscillation of the car, you know that your shock absorbers need changing. The spring on each wheel of a car has a spring constant of \(4005 \mathrm{~N} / \mathrm{m}\), and the car has a mass of \(851 \mathrm{~kg}\), equally distributed over all four wheels. Its shock absorbers have gone bad and provide only \(60.7 \%\) of the damping they were initially designed to provide. What will the period of the underdamped oscillation of this car be if the pushing-down shock absorber test is performed?

A massive object of \(m=5.00 \mathrm{~kg}\) oscillates with simple harmonic motion. Its position as a function of time varies according to the equation \(x(t)=2 \sin ([\pi / 2] t+\pi / 6)\). a) What is the position, velocity, and acceleration of the object at \(t=0 \mathrm{~s}^{2}\) b) What is the kinetic energy of the object as a function of time? c) At which time after \(t=0 \mathrm{~s}\) is the kinetic energy first at a maximum?

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum.) Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3}\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this ascillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x| / \alpha\), where \(\alpha\) is any positive value and \(\gamma\) is a constant

Object \(A\) is four times heavier than object B. Fach object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(A\) is half that of \(B ?\) a) \(T_{A}: T_{\mathrm{g}}=1: 4\) c) \(T_{A}: T_{\mathrm{B}}=2\) : b) \(T_{A}: T_{B}=4: 1\) d) \(T_{A}: T_{B}=1: 2\)

A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.
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