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Chapter 14: Problem 14

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{m},\) it takes the block 2.70 s to travel from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} .\) If the amplitude is doubled, to \(0.180 \mathrm{m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{m}\) to \(x=-0.180 \mathrm{m}\) and (b) from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} ?\)

Short Answer

Expert verified
Both journeys take 2.70 seconds; amplitudes don't alter period.

Step by step solution

01

Understand the problem

The problem presents a block moving in Simple Harmonic Motion (SHM) attached to a spring. We need to determine the time it takes for the block to move between specified positions when the amplitude changes.
02

Recall SHM properties

In SHM, the period of oscillation does not depend on the amplitude. It only depends on the mass and the spring constant. Since the motion from one extreme to the other takes half a period, time for this motion is half the period, which is the same regardless of amplitude.
03

Determine Time from x=0.090m to x=-0.090m

Initially given for a smaller amplitude (0.090m), it takes 2.70 seconds. This time is exactly half the period of the oscillation.
04

Calculate time for double amplitude from x=0.180m to x=-0.180m

Since the period (and hence half the period) does not change with amplitude, it will still take the same amount of time, 2.70 seconds, to travel from one extreme to the other.
05

Calculate time from x=0.090m to x=-0.090m for double amplitude

For this part, with the amplitude doubled, the motion within the portion x = 0.090 m to x = -0.090 m constitutes a smaller journey than the full sweep from 0.180 m to -0.180 m. Yet, it takes the same amount of time (2.70 seconds) since it corresponds to the same change in phase angle as with smaller amplitude.

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Key Concepts

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

Amplitude in SHM
In simple harmonic motion (SHM), the amplitude is a crucial factor as it represents the maximum extent of the object's displacement from its equilibrium position. Imagine stretching a spring and then letting it oscillate back and forth; the amplitude is the farthest point it reaches during these swings.
  • Amplitude is denoted by the letter 'A' and is measured in meters.
  • It does not affect the time period of oscillation in ideal SHM.
  • In real-world situations, amplitude could be influenced by damping forces, but in this frictionless scenario, it remains constant for each oscillation cycle.
Understanding amplitude was essential in our problem because even though the block's amplitude was doubled, it didn't change the time taken for various trips across the same distance, given the properties of SHM.
Period of Oscillation and Its Independence from Amplitude
The period of oscillation in SHM is the time it takes for an oscillating system to complete one full cycle. This period is a fundamental property, determined by the system's mass and the spring constant, not by its amplitude. For example:
  • In the problem, the period of the block's oscillation remains the same whether the amplitude is 0.090 m or 0.180 m.
  • The time for a half-cycle (from one extreme position to the other) is half the period.
  • This highlights the key SHM property that the period is determined solely by intrinsic factors like mass and spring constant.
This property is critical in physics because it means we can predict the behavior of oscillating systems without worrying about changes in amplitude, as demonstrated in our example.
Understanding the Spring Constant
The spring constant, often denoted by the letter 'k', determines the stiffness of a spring. In SHM, it is one of the main factors that dictate the period of oscillation. A higher spring constant indicates a stiffer spring, which affects how quickly it oscillates.Key points about spring constants:
  • Measured in newtons per meter (N/m).
  • Directly influences the period, with the formula for period in a spring-mass system: \( T = 2\pi\sqrt{\frac{m}{k}} \).
  • In this problem, since the spring was ideal, we assumed its spring constant remained unchanged across different amplitudes.
The spring constant plays a vital role in determining the motion's characteristics and ensures that, regardless of amplitude, the period of oscillation remains constant, emphasizing SHM's predictability.

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Most popular questions from this chapter

An unhappy \(0.300-\mathrm{kg}\) rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\) (a) If the constant \(b\) has the value \(0.900 \mathrm{kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 \(\mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) . You twist the part a small amount about this axis and let it go, timing 125 oscillations in 265 s. What is the moment of inertia you want to find?

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is period on the surface of Mars, where \(g=3.71 \mathrm{m} / \mathrm{s}^{2} ?\)

CP A rifle bullet with mass 8.00 \(\mathrm{g}\) and initial horizontal velocity 280 \(\mathrm{m} / \mathrm{s}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 \(\mathrm{cm}\) . After the impact, the block moves in SHM. Calculate the period of this motion.
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