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\(\bullet\bullet\) The velocity of a vertically oscillating mass-spring system is given by \(v=(0.650 \mathrm{~m} / \mathrm{s}) \sin [(4 \mathrm{rad} / \mathrm{s}) t]\) Determine (a) the amplitude and (b) the maximum acceleration of this oscillator.

Short Answer

Expert verified
(a) Amplitude is 0.1625 m. (b) Maximum acceleration is 2.6 m/s².

Step by step solution

01

Identify the General Equation for Velocity

The velocity of the oscillating mass-spring system is given by the equation \( v = A \omega \sin(\omega t) \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. Comparing this with the given equation \( v = (0.650 \text{ m/s}) \sin[(4 \text{ rad/s}) t] \), we identify \( A \omega = 0.650 \text{ m/s} \) and \( \omega = 4 \text{ rad/s}.\)
02

Calculate the Amplitude

To find the amplitude \( A \), use the relation \( A \omega = 0.650 \text{ m/s} \). Substituting \( \omega = 4 \text{ rad/s} \) gives \( A = \frac{0.650}{4} \text{ m} = 0.1625 \text{ m}.\)
03

Find Maximum Acceleration

The maximum acceleration in a mass-spring system is given by \( a_{\text{max}} = A \omega^2 \). Substitute \( A = 0.1625 \text{ m} \) and \( \omega = 4 \text{ rad/s} \) into this formula: \( a_{\text{max}} = 0.1625 \cdot 4^2 = 0.1625 \cdot 16 \text{ m/s}^2 = 2.6 \text{ m/s}^2.\)

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Key Concepts

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

Understanding Oscillating Systems
Oscillating systems are fascinating to study due to their repetitive and predictable nature. These systems include anything that moves back and forth, like the classic mass attached to a spring. When the mass is displaced from its rest position, the spring's force acts to pull it back. This movement causes the mass to oscillate about an equilibrium point. Oscillatory motion can be observed in many natural and mechanical systems, such as a swinging pendulum or vibrating guitar string.

In a mass-spring system, the interplay between kinetic and potential energy results in continuous motion until external forces dampen the system. Key parameters are involved in describing the system's behavior, such as amplitude, angular frequency, and maximum acceleration. These terms provide insight into how fast and how far the system oscillates, making them crucial for understanding and predicting the motion of oscillating systems.
Amplitude Calculation
The amplitude of an oscillating system like the mass-spring setup is the maximum extent of movement from the equilibrium position. It gives us a measure of the system's energy and is closely connected to how far the mass moves up and down during each oscillation.

In the context of our given problem, amplitude is calculated using the equation for velocity of the system:
\[ v = A \omega \sin(\omega t) \] Where "\( A \)" is the amplitude and "\( \omega \)" is the angular frequency. As revealed in the step-by-step solution, we can obtain \( A \) by dividing the given velocity result by the angular frequency. Using \( \omega = 4 \text{ rad/s} \), the amplitude \( A \) is computed as:
  • \( A = \frac{0.650}{4} \text{ m} = 0.1625 \text{ m} \)
The amplitude is crucial because it shows us the scale of oscillation, revealing how broad the oscillation is and how much energy is stored in the oscillatory motion.
Maximum Acceleration in Oscillating Systems
Maximum acceleration in a mass-spring system occurs at the points of maximum displacement, where the spring force is highest. This is when the system experiences the greatest force trying to return it to the equilibrium point.

To find the maximum acceleration, we use the formula:
\[ a_{\text{max}} = A \omega^2 \]
Where \( A \) is the amplitude, and \( \omega \) is the angular frequency. Applying the values calculated previously, with \( A = 0.1625 \text{ m} \) and \( \omega = 4 \text{ rad/s} \), we get:
  • \( a_{\text{max}} = 0.1625 \times 4^2 = 0.1625 \times 16 = 2.6 \text{ m/s}^2 \)
This calculation shows us how quickly the velocity of the mass changes at the most stretched or compressed points of the spring. The max acceleration is pivotal in determining the intensity of the forces acting within the system and helps understand the dynamics of oscillatory motion.
Angular Frequency Explained
Angular frequency \( \omega \) is a fundamental concept in studying oscillations. It represents how many cycles the system completes in a unit of time and is directly related to the speed of the oscillatory motion. Unlike regular frequency measured in hertz (cycles per second), angular frequency is expressed in radians per second. This ties the concept closer to angular physics, which is often used in circular motions.

In the mathematical representation of a mass-spring system, angular frequency appears in the equations for velocity and acceleration:
  • Velocity: \( v = A \omega \sin(\omega t) \)
  • Acceleration: \( a = A \omega^2 \cos(\omega t) \)
Given our exercise data, the angular frequency \( \omega \) was provided as \( 4 \text{ rad/s} \). This means the system completes oscillations rapidly, at a rate of 4 radians per second. Understanding angular frequency is crucial because it dictates how fast the system oscillates, influencing the timing and speed of all motion aspects, which is essential for predicting and modeling real-world oscillatory behaviors.

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Most popular questions from this chapter

\(\bullet\bullet\) A \(\mathrm{} 0.25-\mathrm{kg}\) object is suspended on a light spring of spring constant \(49 \mathrm{~N} / \mathrm{m}\). The spring is then compressed to a position \(15 \mathrm{~cm}\) above the stretched equilibrium position. How much more energy does the system have at the compressed position than at the stretched equilibrium position?

To study the effects of acceleration on the period of oscillation, a student puts a grandfather clock with a \(0.9929-\mathrm{m}-\) long pendulum inside an elevator. Find the period of the grandfather clock (a) when the elevator is stationary, (b) when the elevator is accelerating upward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) (c) when the elevator is accelerating downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2},\) (d) when the cable on the elevator breaks and the elevator simply falls, and (e) when the elevator is moving upward at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\).

Spring \(\mathrm{A}(50.0 \mathrm{~N} / \mathrm{m})\) is attached to the ceiling. The top of spring \(\mathrm{B}(30.0 \mathrm{~N} / \mathrm{m})\) is hooked onto the bottom of spring A. Then a 0.250-kg mass is then attached to the bottom of Spring B. (a) How far will the object fall until it reaches equilibrium? (b) What is the period of the resulting oscillation?

You are setting up two standing string waves. You have a length of uniform piano wire that is 3.0 m long and has a mass of 0.150 kg. You cut this into two lengths, one of 1.0 m and the other of 2.0 m, and place each length under tension. What should be the ratio of tensions (expressed as short to long) so that their fundamental frequencies are the same?

\(\bullet\bullet\) (a) If the mass in a mass-spring system is halved, the new period is \((1) 2,(2) \sqrt{2},(3) 1 / \sqrt{2},(4) 1 / 2\) times the old period. Why? (b) If the initial period is \(3.0 \mathrm{~s}\) and the mass is reduced to \(1 / 3\) of its initial value, what is the new period?

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