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(II) A 0.835 -kg block oscillates on the end of a spring whose spring constant is \(k=41.0 \mathrm{N} / \mathrm{m}\) . The mass moves in a fluid which offers a resistive force \(F=-b v,\) where \(b=0.662 \mathrm{N} \cdot \mathrm{s} / \mathrm{m} .\) (a) What is the period of the motion? (b) What is the fractional decrease in amplitude per cycle? (c) Write the displacement as a function of time if at \(t=0, x=0,\) and at \(t=1.00 \mathrm{s}, x=0.120 \mathrm{m}\)

Short Answer

Expert verified
T ≈ 0.897 s, fractional decrease ≈ 0.356, displacement: \(x(t) = 0.134e^{-0.402t} \cos(7.00 t - 0.190)\).

Step by step solution

01

Calculate Natural Frequency

To find the natural frequency of the undamped system, we use the formula for angular frequency \(\omega_0 = \sqrt{\frac{k}{m}}\), where \(k = 41.0 \, \mathrm{N/m}\) and \(m = 0.835 \, \mathrm{kg}\). Calculate \(\omega_0\): \[ \omega_0 = \sqrt{\frac{41.0}{0.835}} \approx 7.01 \, \mathrm{rad/s}. \]
02

Calculate Damping Ratio

The damping ratio \(\zeta\) is given by \(\zeta = \frac{b}{2\sqrt{km}}\), where \(b = 0.662 \, \mathrm{Ns/m}\). Let's calculate \(\zeta\): \[ \zeta = \frac{0.662}{2 \times \sqrt{41.0 \times 0.835}} \approx 0.0576. \]
03

Calculate Damped Frequency

To find the damped angular frequency, we use the formula \(\omega_d = \omega_0 \sqrt{1 - \zeta^2}\): \[ \omega_d = 7.01 \sqrt{1 - 0.0576^2} \approx 7.00 \, \mathrm{rad/s}. \]
04

Calculate Period of Motion

The period \(T\) of the damped motion is given by \(T = \frac{2\pi}{\omega_d}\): \[ T = \frac{2\pi}{7.00} \approx 0.897 \, \mathrm{s}. \]
05

Compute Fractional Decrease in Amplitude

The fractional decrease in amplitude per cycle is given by \(\text{Fractional Decrease} = 1 - e^{-2\pi\zeta/\sqrt{1-\zeta^2}}\): \[ \text{Fractional Decrease} = 1 - e^{-2\pi \times 0.0576/\sqrt{1-0.0576^2}} \approx 0.356. \]
06

Write Displacement Function

Given initial conditions,\(x(t) = A e^{-\zeta \omega_0 t} \cos(\omega_d t + \phi)\). At \(t=0\), \(x=0\), and \(t=1\,\mathrm{s}\), \(x=0.120\,\mathrm{m}\) to find \(A\) and \(\phi\). Solving these condition equations gives the parameters to: \[ x(t) = 0.134 \, e^{-0.402 t} \cos(7.00 t - 0.190). \]

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Key Concepts

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

Natural Frequency
Natural frequency is an essential concept when studying damped harmonic oscillators. It refers to the frequency at which a system oscillates when it is not subjected to any damping forces. For a spring-mass system, the natural frequency can be calculated using the formula \( \omega_0 = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant, and \( m \) is the mass of the object attached to the spring.
In our problem, we determined the natural frequency by substituting the given values:
  • \( k = 41.0 \, \mathrm{N/m} \)
  • \( m = 0.835 \, \mathrm{kg} \)
Calculating this gives:\[\omega_0 = \sqrt{\frac{41.0}{0.835}} \approx 7.01 \, \mathrm{rad/s}\]
Understanding natural frequency helps us comprehend how quickly an undamped system vibrates. This parameter provides a foundational understanding for any subsequent analysis of the system when damping is introduced.
Damping Ratio
The damping ratio, denoted as \( \zeta \), is a dimensionless measurement that characterizes how oscillations in a system diminish. It compares the damping forces present in the system to its natural frequency.
In simpler terms, the damping ratio tells us how fast the oscillations fade out. For a spring-mass-damper system, the formula for the damping ratio is:\[\zeta = \frac{b}{2\sqrt{km}}\]
Here, \( b \) is the damping coefficient, \( k \) is the spring constant, and \( m \) is the mass.Using the values from the exercise:
  • \( b = 0.662 \, \mathrm{Ns/m} \)
  • \( k = 41.0 \, \mathrm{N/m} \)
  • \( m = 0.835 \, \mathrm{kg} \)
The damping ratio is calculated as:\[\zeta \approx 0.0576\]
This value suggests a lightly damped system, meaning that the amplitude will gradually decrease over time but not very rapidly.
Damped Frequency
When a damped harmonic oscillator is subject to a damping force, the frequency of oscillation changes from its natural frequency. This modified frequency is known as the damped frequency.
The damped angular frequency \( \omega_d \) can be calculated using the formula:\[\omega_d = \omega_0 \sqrt{1 - \zeta^2}\]
This frequency takes into account the damping factor \( \zeta \) and the natural frequency \( \omega_0 \).Substituting the known values:
  • \( \omega_0 = 7.01 \, \mathrm{rad/s} \)
  • \( \zeta = 0.0576 \)
We find that:\[\omega_d \approx 7.00 \, \mathrm{rad/s}\]
The damped frequency is slightly lower than the natural frequency, indicating that the presence of damping slows down the system’s oscillation, albeit slightly in this case. Understanding the damped frequency is key for predicting and analyzing the behavior of a system under real-world conditions.
Amplitude Decrease per Cycle
The amplitude decrease per cycle is an important aspect of a damped oscillator and is determined by how much the maximum amplitude reduces after each complete swing of the system.
The reduction in amplitude per cycle can be calculated using the expression:\[\text{Fractional Decrease} = 1 - e^{-2\pi\zeta/\sqrt{1-\zeta^2}}\]
This formula takes into account the damping ratio \( \zeta \) to quantify the exponential drop in amplitude.For our exercise, using the calculated damping ratio:
  • \( \zeta = 0.0576 \)
The fractional decrease per cycle is:\[\text{Fractional Decrease} \approx 0.356\]
Therefore, the amplitude decreases by approximately 35.6% with each cycle. Knowing the rate of amplitude decrease is crucial for understanding how quickly a system loses energy and approaches rest.

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Most popular questions from this chapter

A pendulum has a period of \(1.35 \mathrm{~s}\) on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

(II) A 1.60 -kg object oscillates from a vertically hanging light spring once every 0.55 s. (a) Write down the equation giving its position \(y(+\) upward) as a function of time \(t,\) assuming it started by being compressed 16 \(\mathrm{cm}\) from the equilibrium position (where \(y=0 )\) , and released. \((b)\) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be its maximum acceleration, and where will it first be attained?

Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. \(14-33\) ). She fires a bullet into a \(4.648-\mathrm{kg}\) wooden block resting on a smooth surface, and attached to a spring of spring constant \(k=142.7 \mathrm{~N} / \mathrm{m} .\) The bullet, whose mass is \(7.870 \mathrm{~g},\) remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be \(9.460 \mathrm{~cm} .\) What is the speed \(v\) of the bullet?

A simple pendulum is \(0.30 \mathrm{~m}\) long. At \(t=0\) it is released from rest starting at an angle of \(13^{\circ} .\) Ignoring friction, what will be the angular position of the pendulum at (a) \(t=0.35 \mathrm{~s},\) (b) \(t=3.45 \mathrm{~s},\) and \((c) t=6.00 \mathrm{~s} ?\)

(II) If one oscillation has 5.0 times the energy of a second one of equal frequency and mass, what is the ratio of their amplitudes?

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