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CE An object undergoes simple harmonic motion with a period T. In the time \(3 T / 2\) the object moves through a total distance of \(12 D .\) In terms of \(D,\) what is the object's amplitude of motion?

Short Answer

Expert verified
The amplitude of motion is \(2D\).

Step by step solution

01

Understand the problem

We are given the object's period of motion as \(T\), and we know that in time \(\frac{3T}{2}\), it moves a total distance of \(12D\). We need to find the amplitude \(A\) of this harmonic motion in terms of \(D\).
02

Calculate fraction of cycles

Since the time interval given is \(\frac{3T}{2}\), we know this represents \(\frac{3}{2}\) of a complete cycle. Simple harmonic motion covers a certain distance in each fraction of its cycle.
03

Calculate distance for full cycle

For simple harmonic motion, one complete cycle is a movement from one extreme to the opposite and back again. This means a round trip of distance: \(4A\).
04

Express total distance in terms of A

Since \(\frac{3}{2}\) of the cycle corresponds to a distance of \(12D\), we can equate \(\frac{3}{2} \times 4A = 12D\).
05

Solve for amplitude A

Solve the equation for \(A\): \[ \frac{3}{2} \times 4A = 12D \] Simplifying gives: \[ 6A = 12D \] Therefore, \[ A = 2D \].

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Key Concepts

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

Amplitude
In the context of simple harmonic motion, amplitude refers to the maximum displacement of an object from its equilibrium or rest position. It can be thought of as the "height" of the wave when visualizing the motion, and it is a key characteristic of oscillatory movement. When an object undergoes simple harmonic motion, like a swinging pendulum, it moves back and forth in a regular pattern. The farthest points that the object reaches on either side of its path define the amplitude. For example:
  • If a pendulum swings 2 meters in both directions from its resting position, the amplitude is 2 meters.
  • The object's motion is symmetrical with respect to the equilibrium, which means it swings equally far in both the positive and negative directions.
In exercises involving simple harmonic motion, you may be given other information, such as the total distance traveled over part of the cycle, to determine the amplitude. It always remains consistent during the whole motion unless influenced by damping forces.
Period
The period of simple harmonic motion is the time it takes to complete one full cycle of motion. It is represented by the symbol \(T\) and is typically measured in seconds. Understanding the period is crucial because it provides information about how quickly the oscillations occur.Here's a breakdown of how the period functions:
  • The period is the duration from the starting point back to the same point after a full oscillation.
  • For instance, consider a pendulum swinging back and forth. If it takes 5 seconds to return to the starting point, its period \(T\) is 5 seconds.
  • The period is inversely related to frequency, which is the number of cycles that occur within a second. Frequency is given by the equation \(f = \frac{1}{T}\).
The period is independent of other factors like amplitude in simple systems. However, it can be influenced by characteristics such as the length of a pendulum or the mass of an object if additional forces are involved. For a system in simple harmonic motion with a period \(T\), the full cycle is completed in that specific time.
Cycle
A cycle in simple harmonic motion refers to the complete path that an oscillating object takes before repeating the pattern again. Characteristics of a cycle in this context include:
  • Starting at an equilibrium point, moving to an extreme on one side (maximum displacement), back through equilibrium, and reaching the maximum displacement on the opposite side before returning to the starting point.
  • Each cycle in simple harmonic motion is defined by two extreme positions and the path between them.
  • A cycle is complete when the motion returns to its starting condition, much like drawing a circle.
The distance moved during one cycle is often four times the amplitude, denoted as \(4A\). This understanding forms the basis for solving many problems involving cycles, such as calculating the total distance traveled over fractions of a cycle. Recognizing how an object behaves in one complete cycle helps in understanding its path at any given moment in its oscillation.

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Most popular questions from this chapter

\(\cdot \cdot\) On the construction site for a new skyscraper, a uniform beam of steel is suspended from one end. If the beam swings back and forth with a period of \(2.00 \mathrm{s},\) what is its length?

A mass oscillates on a spring with a period \(T\) and an amplitude \(0.48 \mathrm{cm}\). The mass is at the equilibrium position \(x=0\) at \(t=0,\) and is moving in the positive direction. Where is the mass at the times (a) \(t=T / 8\) (b) \(t=T / 4\) (c) \(t=T / 2\) and (d) \(t=3 T / 4 ?\) (e) Plot your results for parts (a) through (d) with the vertical axis representing position and the horizontal axis representing time.

A ball rolls on a circular track of radius \(0.62 \mathrm{m}\) with a constant angular speed of \(1.3 \mathrm{rad} / \mathrm{s}\) in the counterclockwise direction. If the angular position of the ball at \(t=0\) is \(\theta=0,\) find the \(x\) component of the ball's position at the times \(2.5 \mathrm{s}, 5.0 \mathrm{s},\) and \(7.5 \mathrm{s}\). Let \(\theta=0\) correspond to the positive \(x\) direction.

IP A \(0.40-k g\) mass is attached to a spring with a force constant of \(26 \mathrm{N} / \mathrm{m}\) and released from rest a distance of \(3.2 \mathrm{cm}\) from the equilibrium position of the spring. (a) Give a strategy that allows you to find the speed of the mass when it is halfway to the equilibrium position. (b) Use your strategy to find this speed.

An object moves with simple harmonic motion of period T and amplitude A. During one complete cycle, for what length of time is the speed of the object greater than \(v_{\max } / 2 ?\)

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