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Chapter 13: Problem 27

You are watching an object that is moving in SHM. When the object is displaced 0.600 \(\mathrm{m}\) to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Short Answer

Expert verified
The object moves an additional 0.240 m to the right before stopping.

Step by step solution

01

Identify the given information

We have the following information:1. Displacement from equilibrium: \(x = 0.600 \, \text{m}\)2. Velocity at this point: \(v = 2.20 \, \text{m/s}\) towards the right3. Acceleration at this point: \(a = -8.40 \, \text{m/s}^2\) (to the left meaning negative)We need to find the amplitude of the Simple Harmonic Motion (SHM) to determine the maximum displacement from the equilibrium.
02

Use the relation between acceleration, displacement, and angular frequency

In Simple Harmonic Motion, the acceleration \(a\) is given by:\[ a = -\omega^2 x \]where \(\omega\) is the angular frequency.Using the given values:\[ -8.40 = -\omega^2 (0.600) \]Solving for \(\omega^2\):\[ \omega^2 = \frac{8.40}{0.600} = 14 \, \text{s}^{-2} \]
03

Relate velocity, displacement and angular frequency

The velocity \(v\) in SHM is given by:\[ v^2 = \omega^2 (A^2 - x^2) \]where \(A\) is the amplitude of the motion.Substitute the known values:\[ (2.20)^2 = 14 (A^2 - (0.600)^2) \]\[ 4.84 = 14 (A^2 - 0.360) \]\[ A^2 - 0.360 = \frac{4.84}{14} \]\[ A^2 - 0.360 = 0.3457 \]\[ A^2 = 0.7057 \]\[ A = \sqrt{0.7057} \approx 0.840 \, \text{m} \]
04

Find the additional displacement

The object is currently at \(x = 0.600 \, \text{m}\) and the amplitude is \(A = 0.840 \, \text{m}\). Therefore, the additional displacement before it stops is:\[ \text{Additional Displacement} = A - x = 0.840 - 0.600 = 0.240 \, \text{m} \]
05

Summarize the result

The object will move an additional \(0.240 \, \text{m}\) to the right before it momentarily stops and changes direction.

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Key Concepts

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

Angular Frequency
Angular frequency is a key concept in understanding Simple Harmonic Motion (SHM). It essentially describes how quickly something oscillates back and forth. Imagine a pendulum swinging or a spring bouncing up and down. Here, angular frequency (\( \omega \)) tells us how fast these cycles repeat over time.
In the case of SHM, the angular frequency is a fundamental property that connects acceleration, displacement, and velocity. The formula \( a = -\omega^2 x \) shows that acceleration is directly related to both the angular frequency and displacement.
The negative sign indicates that acceleration is always directed towards the equilibrium position.
Understanding this relationship helps us determine how the oscillating object will behave under different conditions. It's important to note that larger angular frequencies mean faster oscillations.
  • Angular frequency (\( \omega \)) is measured in radians per second.
  • It indicates the speed of oscillation in SHM.
  • It's determined by the system's properties, like mass and stiffness of a spring.
Displacement
Displacement in Simple Harmonic Motion refers to how far an object is moved from its equilibrium or rest position. It's a vector quantity, meaning it not only has a magnitude (how much) but also a direction.
In our exercise, the object is displaced 0.600 m to the right. Displacement is crucial because it informs us where the object is at any point within its oscillatory movement.
By knowing the displacement and using equations that involve angular frequency and amplitude, such as \( v^2 = \omega^2 (A^2 - x^2) \) , we can predict future positions and velocities.
  • Displacement can be positive or negative, depending on the direction from equilibrium.
  • It's integral in calculating velocity and acceleration at any point.
  • Helps to determine how much further the object can move (like how in the exercise, an additional 0.240 m was calculated).
Amplitude
Amplitude represents the maximum extent of displacement from the equilibrium position in Simple Harmonic Motion. It is a crucial factor because it defines the boundaries of the object's motion; the object will never swing or bounce further than this value from the midpoint.
In the provided problem, the amplitude (\( A \)) was calculated as 0.840 m. This tells us how far the object can move from its center point before reversing direction.
This property is vital for determining how objects will behave in various physical systems using SHM principles. The amplitude is directly linked to the energy in the system, as larger amplitudes generally mean more energy is needed.
  • Amplitude is always positive and is expressed in meters in SI units.
  • It signifies the maximum reach, or half the total distance the motion covers back and forth.
  • Knowing the amplitude allows determination of unknowns like the additional displacement.

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Most popular questions from this chapter

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\Theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\Theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a \(0.200-\mathrm{kg}\) pan is suspended from its lower end. A butcher drops a 2.2\(\cdot \mathrm{kg}\) steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak imto immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

A 0.500 \(\mathrm{kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) . What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 \(\mathrm{s}\) its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

The two pendulums shown in Fig 13.34 each consist of a uniform solid ball of mass \(M\) supported by a massless string, but the ball for pendulum \(A\) is very tiny while the ball for pendulum \(B\) is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?
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