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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 \(1.00 \mathrm{~s} ?\)

Short Answer

Expert verified
Answer: It takes 9.21 seconds for the amplitude to decrease to 1% of its original value. Question: What damping constant is needed to reduce the amplitude by 99% in 1 second? Answer: The damping constant needed to reduce the amplitude by 99% in 1 second is 27.73 kg/s.

Step by step solution

01

Find the decay factor e^(-bt/2m)

Recall that the decay factor for a damped harmonic oscillator is given by \(e^{(-bT / 2m)}\), where \(T\) is the time, \(b\) is the damping constant and \(m\) is the mass. This decay factor represents the percentage of the initial amplitude remaining at time \(T\).
02

Calculate the time T for amplitude to decay to 1%

To find the time it takes for the amplitude to reach 1% of its initial value, we need to solve the equation \(e^{(-10T / (2*3))} = 0.01\). Divide both sides by the mass and rearrange the equation to get: \(T = -\frac{2m\ln{(0.01)}}{b}\). Then, plug in the given values of \(m\) and \(b\) to get the time. \(T = -\frac{2(3)\ln{(0.01)}}{10} = 9.21\) seconds So, it takes 9.21 seconds for the amplitude to decrease to 1% of its original value.
03

Calculate the damping constant b for amplitude to decay to 1% in 1 second

Now, we want to find the damping constant \(b\) required to decrease the amplitude by 99% in 1 second. This means the decay factor should be 0.01 at \(T=1\). Using the decay factor equation, we have: \(0.01 = e^{(-b / 6)}\). Rearrange the equation to solve for \(b\). \(b = -6\ln{(0.01)} = 27.73\) kg/s So, the damping constant needed to reduce the amplitude by 99% in 1 second is 27.73 kg/s.

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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, often represented by the symbol \b\), is a crucial parameter when analyzing the motion of a damped harmonic oscillator. It quantitatively describes the resistance that a medium, such as oil or air, provides against the motion of the object attached to the spring.
When an object oscillates in a resistive medium, the damping constant slows down the object's motion by dissipating its energy, usually in the form of thermal energy. A higher damping constant indicates a stronger resistance, therefore a quicker loss of the oscillator's energy.
For instance, in the exercise, the damping constant of the oil is given as \(10.0\mathrm{~kg/s}\) and is used to determine how long it will take for the amplitude of oscillations to significantly decrease. The value of \(b\) directly affects the time it takes for the oscillation amplitude to decay. In real-world applications, high damping constants are used in systems where rapid stabilization is needed after a disturbance, such as in shock absorbers in vehicles. Conversely, low damping is desired in systems where sustained vibrations are beneficial, such as in the strings of a musical instrument.
Decay Factor
The decay factor is a mathematical expression describing how fast the amplitude of a damped harmonic oscillator decreases over time. Represented as \( e^{(-bT / 2m)} \), where \( T \) is the elapsed time, \( b \) is the damping constant, and \( m \) is the mass of the oscillating object, it reflects the exponential nature of the decay process.
In the steps provided for the solution, the decay factor is used to determine the amount of time needed for the oscillation's amplitude to diminish to a specific percentage of its original value. For example, to calculate the time it takes for the amplitude to drop to 1% of its original value, the decay factor equation is set to 0.01, and the variables are manipulated to solve for time \( T \).
Understanding the decay factor is fundamental for many scientific and engineering fields. It allows the prediction of how different damping constants and masses can influence the rate of energy loss in oscillating systems, which is essential for designing safe and efficient mechanical and structural systems.
Oscillation Amplitude Decay
Oscillation amplitude decay refers to the diminishing magnitude of the oscillations over time in a damped harmonic oscillator. The decay relates to how quickly the peaks of the oscillatory motion decrease, which is governed by the damping constant and the system's characteristics.
In our exercise example, the goal is to determine the time required for the amplitude to decay to just 1% of its initial value. This is indicative of how amplitude decay can be quantified and used to understand the behavior of oscillating systems under the influence of damping.
Real-life examples of oscillation amplitude decay can be seen in pendulums coming to a rest or the steady decline of vibrations in a plucked guitar string. Understanding this concept is essential for tuning systems to optimize performance, such as reducing unwanted vibrations in engines or tuning the response of electronic circuits.

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Most popular questions from this chapter

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.

A 2.00 -kg mass attached to a spring is displaced \(8.00 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.00 \mathrm{~Hz}\) 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 \(2.00 \mathrm{~cm}\) from the equilibrium position?

A vertical spring with a spring constant of \(2.00 \mathrm{~N} / \mathrm{m}\) has a \(0.300-\mathrm{kg}\) mass attached to it, and the mass moves in a medium with a damping constant of \(0.0250 \mathrm{~kg} / \mathrm{s}\). The mass is released from rest at a position \(5.00 \mathrm{~cm}\) from the equilibrium position. How long will it take for the amplitude to decrease to \(2.50 \mathrm{~cm} ?\)

A spring with \(k=12.0 \mathrm{~N} / \mathrm{m}\) has a mass \(m=3.00 \mathrm{~kg}\) attached to its end. The mass is pulled \(+10.0 \mathrm{~cm}\) from the equilibrium position and released from rest. What is the velocity of the mass as it passes the equilibrium position? a) \(-0.125 \mathrm{~m} / \mathrm{s}\) b) \(+0.750 \mathrm{~m} / \mathrm{s}\) c) \(-0.200 \mathrm{~m} / \mathrm{s}\) d) \(+0.500 \mathrm{~m} / \mathrm{s}\) e) \(-0.633 \mathrm{~m} / \mathrm{s}\)

A 100 .-g block hangs from a spring with \(k=5.00 \mathrm{~N} / \mathrm{m}\). At \(t=0 \mathrm{~s}\), the block is \(20.0 \mathrm{~cm}\) below the equilibrium position and moving upward with a speed of \(200 . \mathrm{cm} / \mathrm{s}\). What is the block's speed when the displacement from equilibrium is \(30.0 \mathrm{~cm} ?\)

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