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Chapter 12: Problem 19

A small object is attached to a horizontal spring, pulled to position \(x=-A\), and released. In one full cycle of its motion, the total distance traveled by the object is A. \(A / 2\). D. \(2 A\). B. \(A / 4\). E. \(4 A\). C. \(A\).

Short Answer

Expert verified
The total distance travelled by the object in one full cycle is \(4 A\) units.

Step by step solution

01

Visualize the problem

Identify the initial position of the object, which is \(x=-A\). This means the object is pulled \(A\) units to the left from the equilibrium point. In a cycle, the object would then move to the right to the extent of \(A\) units (that is, going from \(x=-A\) to \(x=A\)), and then return back to its original position \(x=-A\).
02

Calculate the lengths of each segments of the journey

The distance covered when the object moved from \(x=-A\) to \(x=A\) is \(A - (-A) = 2A\) units. The distance covered when the object returns back to its original position from \(x=A\) to \(x=-A\) is also the same, which is \(2A\) units
03

Compute the total distance travelled in one full cycle

To determine the total distance travelled in one full cycle, we add the distances from both segments of the journey. That is \(2A + 2A = 4A\) units

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Key Concepts

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

Spring-Mass System
In a spring-mass system, an object is attached to a spring, which can compress or stretch. The spring tends to return to its natural length due to elastic force. This kind of system often exhibits simple harmonic motion, meaning it moves back and forth in a regular pattern.
When the object is displaced and released, it experiences restoring force due to the spring, which accelerates it back towards the equilibrium position 鈥� the central point where the spring is neither compressed nor extended.

Key characteristics of spring-mass systems include:
  • The force exerted by the spring is proportional to the displacement (Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement).
  • The motion is periodic, meaning it repeats in equal intervals of time.
  • The maximum displacement from the equilibrium is called amplitude (\( A \)).
Oscillation Cycle
An oscillation cycle refers to one complete back-and-forth movement of the object in the spring-mass system. To fully grasp this, picture the object being initially pulled to its maximum displacement (\( -A \)), released, and allowed to travel to the opposite side (+\( A \)). Then, it retraces its path back to the starting point.

The journey spans the following steps:
  • From two negative extreme (\( x = -A \)) to equilibrium (\( x = 0 \)).
  • Then to the positive extreme (\( x = A \)).
  • Back through equilibrium (\( x = 0 \)) to the starting point (\( x = -A \)).

This complete back-and-forth motion defines the 'cycle,' making it easy to assess how distance and timing relate in repetitive systems like these.
Total Distance Traveled
The total distance traveled in one oscillation cycle of a spring-mass system can sometimes be confusing. It doesn't simply stop at the farthest point away from the start. Instead, the object continues its travel back to the starting position. In essence, it travels back and forth fully.

Let's break it down:
  • From the starting position (\(-A\)) to the far end (\(A\)), the distance is \(2A\).
  • From \(A\) back to \(-A\), it covers another \(2A\).

Adding these gives the total distance of \(4A\). The object effectively travels twice across its total range, which solidifies the understanding that the whole cycle encompasses a full back-and-forth journey.

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Most popular questions from this chapter

A \(20.0\)-cm string is made by combining ten parallel cylindrical strands of human hair, each \(20.0 \mathrm{~cm}\) long. Young's modulus for the hair is \(4.50 \mathrm{GN} / \mathrm{m}^{2}\), and each strand is \(125 \mu \mathrm{m}\) thick. A \(175-\mathrm{g}\) utensil is tied to one end of the string and the other end is fastened to the ceiling. (a) By how much does the string stretch beyond its original \(20.0 \mathrm{~cm}\) length when the utensil is attached? (b) If the utensil is now pulled down an additional \(1.40 \mathrm{~mm}\) and then released, how long will it take for the utensil to first return to the position from which it was released? SSM

(a) What is the period of a simple pendulum of length \(1 \mathrm{~m}\) at the top of Mt. Everest, \(8848 \mathrm{~m}\) above sea level. (b) Express your answer as a number times \(T_{0}\), the period at sea level where \(h\) equals 0 . The acceleration due to gravity in terms of elevation is $$ g=g_{0}\left(\frac{R_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}+h}\right)^{2} $$ where \(g_{0}\) is the average acceleration due to gravity at sea level, \(R_{\mathrm{E}}\) is Earth's radius, and \(h\) is elevation above sea level. Take \(g_{0}\) to be \(9.800 \mathrm{~m} / \mathrm{s}^{2}\) and Earth's radius to be \(R_{\mathrm{E}}\) is \(6.380 \times 10^{6} \mathrm{~m}\).

Explain the difference between a simple pendulum and a physical pendulum.

An object on the end of a spring oscillates with a frequency of \(15 \mathrm{~Hz}\). Calculate (a) the period of the motion and (b) the number of oscillations that the object undergoes in \(2 \mathrm{~min}\). SSM

The acceleration of an object that has a mass of \(0.025 \mathrm{~kg}\) and exhibits simple harmonic motion is given by \(a(t)=\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) \cos (\pi t+\pi / 2)\). Calculate its velocity at \(t=2 \mathrm{~s}\), assuming the object starts from rest at \(t=0\). SSM
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