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(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?

Short Answer

Expert verified
The object takes 0.137 seconds to reach equilibrium, with a maximum speed of 1.83 m/s, and a maximum acceleration of 20.79 m/s² at the start.

Step by step solution

01

Determine the angular frequency

The angular frequency \( \omega \) can be calculated using the formula: \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of oscillation. Substituting \( T = 0.55 \text{ s} \), we have: \( \omega = \frac{2\pi}{0.55} \approx 11.42 \text{ rad/s} \).
02

Write the position function

The position function for the simple harmonic motion is \( y(t) = A\cos(\omega t + \phi) \). Since the object starts from the compressed position (\( y = -0.16 \text{ m} \) when \( t = 0 \)), we set \( \phi = \pi \). Thus, the equation becomes \( y(t) = 0.16\cos(11.42t + \pi) = -0.16\cos(11.42t) \).
03

Find the time to reach equilibrium position

The object reaches the equilibrium position \( y = 0 \) when \( cos(11.42t) = 0 \). Solving \( 11.42t = \frac{\pi}{2} \) gives \( t = \frac{\pi}{2 \times 11.42} \approx 0.137 \text{ s} \).
04

Calculate maximum speed

The maximum speed \( v_{max} \) in simple harmonic motion is given by \( v_{max} = A\omega \). Substituting the values, \( v_{max} = 0.16 \times 11.42 \approx 1.83 \text{ m/s} \).
05

Calculate maximum acceleration and its position

The maximum acceleration \( a_{max} \) is given by \( a_{max} = A\omega^2 \). Substituting the values, \( a_{max} = 0.16 \times (11.42)^2 \approx 20.79 \text{ m/s}^2 \). Maximum acceleration occurs at the amplitude, which is first attained when the object is released.

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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. It's a measure of how fast an object oscillates in a cycle as it moves back and forth from its equilibrium position. Think of it as how quickly each cycle of the motion completes.
To calculate angular frequency (\omega), you use the formula:
  • \( \omega = \frac{2\pi}{T} \)
where \( T \) is the period of the oscillation. The period is the time it takes for a complete cycle.
In our example, the period is 0.55 seconds, leading to an angular frequency of about 11.42 radians per second. This tells us that the object completes a cycle in a little over half a second.
Equilibrium position
The equilibrium position is where the forces acting on the object are balanced, hence the object finds itself at rest if undisturbed. In simple harmonic motion, this position refers to the central point where oscillation occurs.
When discussing an oscillating system, equilibrium can either be vertical or horizontal, dependent on the system orientation. Our example uses the vertical orientation, meaning equilibrium is where \(y = 0\).
  • This position is significant because it's the point where the object experiences the highest speed.
  • The object reaches this point quickly, as calculated in the solution - approximately 0.137 seconds from a released position.
Understanding this concept clarifies why systems using simple harmonic motion naturally return to their equilibrium position over time.
Maximum speed
Maximum speed in simple harmonic motion occurs as the object passes through the equilibrium position. At this point, all potential energy converts into kinetic, marking the fastest speed possible during oscillation. The formula to determine maximum speed (\(v_{max}\)) is:
  • \(v_{max} = A\omega\)
where \(A\) is the amplitude and \(\omega\) is the angular frequency.
In the given example, with an amplitude of 0.16 meters and an angular frequency of about 11.42 radians per second, the maximum speed is calculated to be around 1.83 m/s. Recognizing where and why maximum speed occurs helps understand energy conversion in physics.
Maximum acceleration
Maximum acceleration in simple harmonic motion shows how swiftly the velocity of an oscillating object changes. It is greatest at the extreme positions of the object's path, where the speed is zero and potential energy is maximum. This is when the force on the object is at its peak, trying to pull or push it back toward equilibrium. The formula for maximum acceleration (\(a_{max}\)) is:
  • \(a_{max} = A\omega^2 \)
where \(A\) is the amplitude and \(\omega\) is the angular frequency.
In our case, with an amplitude of 0.16 meters and angular frequency 11.42 rad/s, maximum acceleration calculates to approximately 20.79 m/s². Recognizing these values assists in predicting behavior and understanding the forces involved in oscillatory motions.

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Most popular questions from this chapter

(II) At \(t=0,\) a \(785-\mathrm{g}\) mass at rest on the end of a horizontal spring \((k=184 \mathrm{N} / \mathrm{m})\) is struck by a hammer which gives it an initial speed of 2.26 \(\mathrm{m} / \mathrm{s} .\) Determine \((a)\) the period and frequency of the motion, \((b)\) the amplitude, (c) the maximum acceleration, (d) the position as a function of time, \((e)\) the total energy, and \((f)\) the kinetic energy when \(x=0.40 A\) where \(A\) is the amplitude.

(II) A damped harmonic oscillator loses 6.0\(\%\) of its mechanical energy per cycle. \((a)\) By what percentage does its frequency differ from the natural frequency \(f_{0}=(1 / 2 \pi) \sqrt{k / m} ?\) (b) After how many periods will the amplitude have decreased to 1\(/ e\) of its original value?

Imagine that a 10 -cm-diameter circular hole was drilled all the way through the center of the Earth (Fig, \(44 ) .\) At one end of the hole, you drop an apple into the hole. Show that, if you assume that the Earth has a constant density, the subsquent motion of the apple is simple harmonic. How long will the apple take to return? Assume that we can ignore all frictional effects.

(II) Estimate the stiffness of the spring in a child's pogo stick if the child has a mass of 35 \(\mathrm{kg}\) and bounces once every 2.0 seconds.

A small fly of mass \(0.25 \mathrm{~g}\) is caught in a spider's web. The web oscillates predominately with a frequency of \(4.0 \mathrm{~Hz}\). (a) What is the value of the effective spring stiffness constant \(k\) for the web? ( \(b\) ) At what frequency would you expect the web to oscillate if an insect of mass \(0.50 \mathrm{~g}\) were trapped?

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