A \(3.00-\mathrm{kg}\) mass attached to a spring with \(k=140 . \mathrm{N} / \mathrm{m}\) is oscillating in a vat of oil, which damps the oscillations. a) If the damping constant of the oil is \(b=10.0 \mathrm{~kg} / \mathrm{s}\), how long will it take the amplitude of the oscillations to decrease to \(1.00 \%\) of its original value? b) What should the damping constant be to reduce the amplitude of the oscillations by \(99.0 \%\) in 1.00 s?

Short Answer

Expert verified
What should the damping constant be in order to reduce the amplitude by 99% in 1.00 second? Answer: It takes approximately 6.91 seconds for the amplitude to decrease to 1% of its original value, and the damping constant should be approximately 27.7 kg/s to reduce the amplitude by 99% in 1.00 second.

Step by step solution

01

Identify the given information

We are given the mass, \(m = 3.00\,\text{kg}\), the damping constant \(b = 10.0\,\text{kg/s}\), and we need to find the time \(t\) for the amplitude to become 1% of its initial value.
02

Set up the amplitude equation

Write down the equation for the amplitude of a damped harmonic oscillator: \(A(t) = A_0e^{-\frac{b}{2m}t}\) We want to find the time \(t\) when \(A(t) = 0.01A_0\). Substitute this into the equation: \(0.01A_0 = A_0e^{-\frac{b}{2m}t}\)
03

Solve for the time t

Divide both sides by \(A_0\) and take the natural logarithm of both sides: \(\ln{0.01} = -\frac{b}{2m}t\) Now, solve for \(t\). We have \(b = 10.0\,\text{kg/s}\) and \(m = 3.00\,\text{kg}\): \(t = -\frac{2m}{b}\ln{0.01} = \frac{-2\cdot3.00\,\text{kg}}{10.0\,\text{kg/s}}\ln{0.01}\) \(t \approx 6.91\,\text{s}\) The amplitude of the oscillations will decrease to 1% of its original value in approximately 6.91 seconds. #b) Find the damping constant to reduce amplitude by 99% in 1.00 s#
04

Set up the equation

We want to find the damping constant \(b\) when the amplitude reduces by 99% in 1.00 s. This means \(A(t) = 0.01A_0\) at \(t = 1.00\,\text{s}\). Write down the amplitude equation: \(A(t) = A_0e^{-\frac{b}{2m}t}\) Substitute the known values: \(0.01A_0 = A_0e^{-\frac{b}{2m}(1.00\,\text{s})}\)
05

Solve for the damping constant b

Divide both sides by \(A_0\) and take the natural logarithm of both sides: \(\ln{0.01} = -\frac{b}{2m}(1.00\,\text{s})\) Now, solve for \(b\). We have \(m = 3.00\,\text{kg}\): \(b = -\frac{2m}{1.00\,\text{s}}\ln{0.01} = -\frac{2\cdot3.00\,\text{kg}}{1.00\,\text{s}}\ln{0.01}\) \(b \approx 27.7\,\text{kg/s}\) The damping constant should be approximately 27.7 kg/s to reduce the amplitude of the oscillations by 99% in 1.00 second.

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Key Concepts

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

Damping Constant
The damping constant, represented often by the symbol 'b', is a critical parameter in the study of damped harmonic oscillators. It quantifies the resistance that a medium (like oil, air, or water) provides against the motion of the oscillator. In simpler terms, it tells us how quickly the oscillations will die out due to the medium's opposition.
When dealing with a damped harmonic oscillator, the damping constant can be understood as a 'frictional' force, but it's important to note that this force always acts against the direction of the object’s velocity, slowing it down. The larger the damping constant, the more rapidly the amplitude of the oscillation decreases over time, leading to what we describe as overdamped, critically damped, or underdamped systems, each showing different behaviors in response to the degree of damping.
Furthermore, when we improve the exercise as provided, identifying the damping constant allows students to understand how altering this value can directly affect the time it takes for the system's amplitude to reach a certain percentage of its initial value. Understanding how to calculate the damping constant is essential for predicting the behavior of various mechanical and electrical systems subjected to damping forces.
Amplitude Decay
Amplitude decay is the process by which the amplitude of an oscillating system decreases over time. It's most commonly observed in damped harmonic oscillators where non-conservative forces, such as the resistance provided by a damping medium, work to reduce the energy of the system.
In the amplitude decay process, the higher the damping constant, the quicker the energy is lost, leading to a faster decay of the oscillation amplitude. The equation describing this decay is exponential in nature and is heavily determined by the damping constant, as demonstrated in the exercise. The precise rate at which the amplitude decays is crucial for many practical applications: from the design of car suspensions and building shock absorbers to the development of electronic filters.
In improving our exercise, clarifying how the exponential function representing amplitude decay is related to time allows students to more accurately graph the decline of amplitude and connect the decay rate with the physics behind energy dissipation in oscillating systems.
Logarithmic Equations
Logarithmic equations are indispensable when solving for variables involved in the exponential decay of an oscillation's amplitude. They allow the inverse process of exponentiation to be performed, such as when isolating the time variable 't' in our exercise. A logarithm answers the question: to what exponent do we need to raise a base (in natural logarithms, the base is the number 'e' approximated to 2.71828) to obtain a certain number?
By using logarithms, we can rewrite exponential decay equations to solve for unknowns, which is exactly what we did to find both the time it takes for the amplitude to decay and for calculating the damping constant required to achieve a specific decay within a set time frame. It's important to explain in the simplest terms how to take the logarithm of both sides of the equation in order to make the variable of interest the subject. This process is a powerful mathematical tool that students will encounter in diverse disciplines, ranging from physics to financial mathematics, where growth or decay processes are modeled.

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Most popular questions from this chapter

A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

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