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Chapter 14: Problem 24

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 \(\mathrm{m}\) and the period is 3.20 \(\mathrm{s} .\) What are the speed and acceleration of the block when \(x=0.160 \mathrm{m} ?\)

Short Answer

Expert verified
The speed is approximately 0.377 m/s, and the acceleration is approximately -0.615 m/s虏 at x = 0.160 m.

Step by step solution

01

Understand the Given and Find the Required Quantities

The problem provides the amplitude of motion as \(A = 0.250\,\mathrm{m}\) and the period of the motion as \(T = 3.20\,\mathrm{s}\). We need to calculate the speed and acceleration of the block when its displacement \(x = 0.160\,\mathrm{m}\).
02

Calculate the Angular Frequency

The angular frequency \(\omega\) is calculated using the formula \(\omega = \frac{2\pi}{T}\). Substituting the given period \(T = 3.20\,\mathrm{s}\):\[\omega = \frac{2\pi}{3.20} \approx 1.96\,\mathrm{rad/s}.\]
03

Determine the Speed at the Desired Displacement

The speed \(v\) in simple harmonic motion is given by the formula:\[v = \omega \sqrt{A^2 - x^2}\]Substituting \(\omega = 1.96\,\mathrm{rad/s}\), \(A = 0.250\,\mathrm{m}\), and \(x = 0.160\,\mathrm{m}\):\[v = 1.96 \times \sqrt{0.250^2 - 0.160^2} \]\[v = 1.96 \times \sqrt{0.0625 - 0.0256} \]\[v = 1.96 \times \sqrt{0.0369} \]\[v = 1.96 \times 0.192 \]\[v \approx 0.377\,\mathrm{m/s}.\]
04

Determine the Acceleration at the Desired Displacement

The acceleration \(a\) in simple harmonic motion is calculated with the formula:\[a = -\omega^2 x\]Substitute \(\omega = 1.96\,\mathrm{rad/s}\) and \(x = 0.160\,\mathrm{m}\):\[a = -(1.96)^2 \times 0.160 \]\[a = -3.8416 \times 0.160 \]\[a \approx -0.615\,\mathrm{m/s^2}.\]

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Key Concepts

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

Amplitude
Amplitude is a fundamental concept in simple harmonic motion (SHM). It represents the maximum extent of an oscillation or how far the object moves away from its central position. In our exercise, the amplitude is given as 0.250 meters. This means the block moves as far as 0.250 meters from the equilibrium point on either side.
Understanding amplitude is important because it helps to define the size of the oscillation.
  • Amplitude is always a positive value.
  • It is one of the basic parameters that describe the motion completely.
In SHM, amplitude does not change over time unless external forces like damping or driving forces are applied. In perfect conditions, like a frictionless surface in this problem, it remains constant.
Angular Frequency
Angular frequency, denoted by \(\omega\), plays a crucial role in describing how fast the object oscillates in SHM. It is closely related to the period \(T\), which is the time taken for one complete cycle of motion.
  • It is measured in radians per second.
  • Calculated using the formula \(\omega = \frac{2\pi}{T}\).
For our particular problem, we calculated the angular frequency as approximately 1.96 rad/s. This value indicates the rate at which the phase of the object changes as it moves through its cycle. A higher angular frequency means the object oscillates more rapidly.
Speed in SHM
The speed of an object in simple harmonic motion changes continuously but is maximum at the equilibrium position and zero at the amplitude. The speed \(v\) at any given point can be determined using the formula \(v = \omega \sqrt{A^2 - x^2}\).
In our exercise, when the block is at \(x = 0.160 \mathrm{m}\), the speed is calculated to be approximately 0.377 m/s.
  • The formula covers how speed depends on both angular frequency and amplitude.
  • Speed decreases as displacement \(x\) moves towards maximum amplitude.
Understanding the speed variation is important as it helps you predict the object's behavior at any point in its motion.
Acceleration in SHM
Acceleration in simple harmonic motion is a crucial aspect defining how the object changes its velocity over time. It is given by the formula \(a = -\omega^2 x\), which shows it is proportional to both the angular frequency and displacement. The negative sign indicates that acceleration opposes the displacement, striving to return the object to equilibrium.
In this exercise, when \(x = 0.160 \mathrm{m}\), the calculated acceleration is approximately \(-0.615 \mathrm{m/s^2}\).
  • Always points towards the equilibrium position.
  • Its magnitude is largest at maximum amplitude and smallest (zero) at equilibrium.
Understanding acceleration in SHM helps explain why the object moves back and forth and gives insights into its dynamics.

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Most popular questions from this chapter

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

Two pendulums have the same dimensions (length \(L )\) and total mass \((m) .\) Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B\) , half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

BIO Weighing Astronauts. This procedure has actually been used to "weigh"" astronauts in space. A 42.5 -kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back- and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed?

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."
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