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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

Two children are on adjacent playground swings of the same height. They are given pushes by an adult and then left to swing. Assuming that each child on a swing can be treated as a simple pendulum and that friction is negligible, which child takes the longer time for one complete swing (has a longer period)? a) the bigger child d) the child given the b) the lighter child biggest push c) neither child

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 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?

A horizontal tree branch is directly above another horizontal tree branch. The elevation of the higher branch is \(9.65 \mathrm{~m}\) above the ground, and the elevation of the lower branch is \(5.99 \mathrm{~m}\) above the ground. Some children decide to use the two branches to hold a tire swing. One end of the tire swing's rope is tied to the higher tree branch so that the bottom of the tire swing is \(0.47 \mathrm{~m}\) above the ground. This swing is thus a restricted pendulum. Start. ing with the complete length of the rope at an initial angle of \(14.2^{\circ}\) with respect to the vertical, how long does it take a child of mass \(29.9 \mathrm{~kg}\) to complete one swing back and forth?

A Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?

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