/*! 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 44 A damped pendulum has a period o... [FREE SOLUTION] | 91Ó°ÊÓ

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A damped pendulum has a period of 0.66 s and a time constant of 4.1 s. How many oscillations will this pendulum make before its amplitude has decreased to \(20 \%\) of its initial amplitude?

Short Answer

Expert verified
The pendulum will make 9 oscillations before its amplitude decreases to 20% of its initial value.

Step by step solution

01

Understand the relationship between amplitude and time constant

The amplitude of a damped oscillator decreases by a factor of \(e\) (approximately 2.71828) every time constant. Therefore, the amplitude of the pendulum after \(t\) seconds is \(A_0 e^{-t/\tau}\), where \(A_0\) is the initial amplitude and \(\tau\) is the time constant.
02

Calculate the time taken for amplitude to reduce to 20%

Now, set this equal to \(0.2A_0\) (since we are asked when the amplitude has decreased to 20% of its initial value) and solve for \(t\). So, \(0.2A_0 = A_0 e^{-t/\tau}\). This simplifies to \(0.2 = e^{-t/\tau}\). Taking the natural logarithm of both sides, we get \(\ln(0.2) = -t/\tau\). So, \(t = -\tau \ln(0.2)\). If we plug in the given time constant \(\tau = 4.1s\), we find that \(t = -4.1 \ln(0.2) \approx 6.03s\).
03

Compute the number of oscillations

To find out the amount of oscillations, divide this time by the period of the pendulum (\(T = 0.66s\)): number of oscillations \(= t/T = 6.03/0.66 \approx 9.14\). Since a pendulum can't complete a fractional oscillation, we must round this off to the nearest whole number, resulting in 9 oscillations.

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Key Concepts

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

Amplitude Decay
The phenomenon of amplitude decay is a key characteristic of a damped pendulum. Imagine a child on a swing - each swing gets a little shorter until the swing eventually comes to a stop. This reduction in swing or amplitude over time is what we refer to as amplitude decay. In physics, this process is described mathematically by the expression \( A(t) = A_0 e^{-t/\tau} \) where \( A(t) \) represents the amplitude after time \( t \) seconds, \( A_0 \) is the original amplitude, and \( \tau \) is what we call the time constant, an important concept which we'll delve into shortly.

In the case of the damped pendulum in our exercise, the goal is to determine when the amplitude decays to just \(20\% \) of \( A_0 \) - a significant drop from its initial height. It's like saying the child's swing height reduced to \(1/5\)th of what it was when they started swinging. Understanding this decay helps us foresee how quickly the pendulum stops making useful swings, or in other scientific applications, how quickly an oscillating system loses its energy.
Time Constant
The time constant, denoted by \( \tau \) and measured in seconds, is a critical value that tells us how swiftly the amplitude decay occurs. It is the time it takes for the amplitude to reduce to \(1/e \) (about \(36.79\% \) of its initial value), where \( e \) is the base of natural logarithms, roughly \(2.71828\). The larger the time constant, the slower the amplitude decays, much like a thick syrup dripping from a spoon versus water - the syrup having a 'larger time constant'.

For the damped pendulum discussed, the time constant is \(4.1\) seconds. This tells us that after roughly \(4.1\) seconds, the swing's amplitude will be a little over one-third of what it started with. Now, when calculating the time \( t \) at which the amplitude falls to \(20\%\) of the original, the time constant helps us find that this occurs around \(6.03\) seconds by using the natural logarithm in our formula.
Oscillation Period
The oscillation period, often simply called the period, is the time it takes for one complete cycle of oscillation. For the pendulum, this is the time for it to swing from one side, all the way to the other, and back again. In our example, the period is \(0.66\) seconds. This is akin to the rhythm of a heartbeat or the tempo of music - it's essential for understanding the timing of oscillations.

To find out how many oscillations the pendulum makes before its amplitude reduces to \(20\% \) of its starting value, we divide the calculated time before this amplitude decay occurs (\(6.03\) seconds) by the period (\(0.66\) seconds). The result is about \(9.14\) oscillations. However, since a pendulum can't complete a partial oscillation, we round to \(9\) full swings. The connection between the time constant and the period gives us a comprehensive picture of both how often the pendulum swings and how quickly it slows down.

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Most popular questions from this chapter

A physics department has a Foucault pendulum, a longperiod pendulum suspended from the ceiling. The pendulum has an electric circuit that keeps it oscillating with a constant amplitude. When the circuit is turned off, the oscillation amplitude decreases by \(50 \%\) in 22 minutes. What is the pendulum's time constant? How much additional time elapses before the amplitude decreases to \(25 \%\) of its initial value?

In a science museum, you may have seen a Foucault pendulum, which is used to demonstrate the rotation of the earth. In one museum's pendulum, the \(110 \mathrm{kg}\) bob swings from a 15.8-m-long cable with an amplitude of \(5.0^{\circ}\). a. What is the period of this pendulum? b. What is the bob's maximum speed? c. What is the pendulum's maximum kinetic energy? d. When the bob is at its maximum displacement, how much higher is it than when it is at its equilibrium position?

The pendulum on a grandfather clock has a period of 2.00 s. If the clock is not wound, the pendulum's amplitude begins to decay at a rate of \(0.53 \%\) each pendulum period. a. What is the time constant of this pendulum? b. What percentage of the pendulum's energy is lost each period?

In taking your pulse, you count 75 heartbeats in 1 min. What are the period (in s) and frequency (in \(\mathrm{Hz}\) ) of your heart's oscillations?

As we've seen, astronauts measure their mass by measuring the period of oscillation when sitting in a chair connected to a spring. The Body Mass Measurement Device on Skylab, a 1970 s space station, had a spring constant of \(606 \mathrm{N} / \mathrm{m}\). The empty chair oscillated with a period of 0.901 s. What is the mass of an astronaut who oscillates with a period of 2.09 s when sitting in the chair?

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