/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 Calculate the ratio of the kinet... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is half its amplitude.

Short Answer

Expert verified
The ratio of kinetic energy to potential energy is 3:1.

Step by step solution

01

Understanding Kinetic and Potential Energy in SHM

For a simple harmonic oscillator with amplitude \( A \), the potential energy at displacement \( x \) is given by \( PE = \frac{1}{2}kx^2 \), where \( k \) is the spring constant. The kinetic energy is \( KE = \frac{1}{2}k(A^2 - x^2) \) as it's the energy when the spring is not at its equilibrium position.
02

Substitute Displacement into Potential Energy Formula

Since the displacement \( x \) is half the amplitude, we have \( x = \frac{A}{2} \). Substitute \( x = \frac{A}{2} \) into the potential energy formula: \[ PE = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{2}k\frac{A^2}{4} = \frac{1}{8}kA^2. \]
03

Substitute Displacement into Kinetic Energy Formula

Using \( x = \frac{A}{2} \), substitute into the kinetic energy formula: \[ KE = \frac{1}{2}k(A^2 - (\frac{A}{2})^2) = \frac{1}{2}k\left(A^2 - \frac{A^2}{4}\right) = \frac{1}{2}k\frac{3A^2}{4} = \frac{3}{8}kA^2. \]
04

Calculate the Ratio of Kinetic Energy to Potential Energy

The ratio of kinetic energy to potential energy is the fraction \( \frac{KE}{PE} \). Using the expressions from the previous steps, calculate this ratio: \[ \frac{KE}{PE} = \frac{\frac{3}{8}kA^2}{\frac{1}{8}kA^2} = \frac{3}{1} = 3. \]

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Key Concepts

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

Kinetic Energy in Simple Harmonic Motion
Kinetic energy (KE) in a simple harmonic oscillator is the energy that depends on the motion of the oscillator. When the system is moving, it means the kinetic energy is at play. For oscillators, this energy becomes significant because it's strongly tied to displacement. The formula for kinetic energy in this context is given by:
  • \( KE = \frac{1}{2} k \left( A^2 - x^2 \right) \)
  • Here, \( k \) is the spring constant, \( A \) is the amplitude, and \( x \) is the displacement from equilibrium.
When the displacement is half the amplitude, it specifically illustrates how the kinetic energy isn't just about speed but how far from rest the object is. With lesser displacement (as in this problem when \( x = \frac{A}{2} \)), it means the object will have covered less distance for its movement. Thus, the kinetic energy here is a reflection of both how far and with what force the object moves.
Potential Energy in Simple Harmonic Motion
Potential energy (PE) is the stored energy depending on the position of the object in a system. For a spring-based harmonic oscillator, this reflects how much the spring is compressed or stretched from its equilibrium position. The potential energy is expressed as:
  • \( PE = \frac{1}{2} k x^2 \)
  • Given \( k \) is the spring constant, and \( x \) is the displacement.
When a system's displacement (\( x \)) is known, we can easily find its potential energy by substituting in the respective values. If you set \( x \) equal to half the amplitude (\( A = \frac{A}{2} \)), potential energy neatly calculates how much energy is concentrated in the position rather than motion. It helps depict how energy shifts back and forth between kinetic and potential forms as the object moves back and forth in the oscillatory pattern.
Understanding the Spring Constant
The spring constant (\( k \)) is a measure of the stiffness of a spring. It defines the relationship between the force needed to extend or compress the spring and the displacement from its rest position. In the context of simple harmonic oscillators, it helps predict how much force the spring will exert back when it is displaced, making it a foundational component of Hooke's Law:
  • \( F = -kx \)
  • The negative sign indicates the force exerted by the spring is always in the opposite direction of displacement.
In oscillatory systems, the spring constant helps dictate both the potential energy and the kinetic energy calculations by influencing how much the spring will try to return to its original position when displaced. A larger spring constant signifies a stiffer spring that requires more force to displace, altering the potential and kinetic energy relationship.

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Most popular questions from this chapter

The pistons in an internal combustion engine undergo a motion that is approximately simple harmonic. If the amplitude of motion is \(3.5 \mathrm{cm},\) and the engine runs at 1700 rev \(/ \mathrm{min}\), find (a) the maximum acceleration of the pistons and (b) their maximum speed.

A child rocks back and forth on a porch swing with an amplitude of \(0.204 \mathrm{m}\) and a period of \(2.80 \mathrm{s}\). Assuming the motion is approximately simple harmonic, find the child's maximum speed.

\(\cdot\) BIO Measuring an Astronaut's Mass An astronaut uses a Body Mass Measurement Device (BMMD) to determine her mass. What is the astronaut's mass, given that the force constant of the BMMD is \(2600 \mathrm{N} / \mathrm{m}\) and the period of oscillation is \(0.85 \mathrm{s} ?\) (See the discussion on page 427 for more details on the BMMD.)

IP Consider a meterstick that oscillates back and forth about a pivot point at one of its ends, (a) Is the period of a simple pendulum of length \(L=1.00 \mathrm{m}\) greater than, less than, or the same as the period of the meterstick? Explain. (b) Find the length \(L\) of a simple pendulum that has a period equal to the period of the meterstick.

ce Predict/Explain A grandfather clock keeps correct time at sea level. If the clock is taken to the top of a nearby mountain, (a) would you expect it to keep correct time, run slow, or run fast? (b) Choose the best explanation from among the following: I. Gravity is weaker at the top of the mountain, leading to a greater period of oscillation. II. The length of the pendulum is unchanged, and therefore its period remains the same. III. The extra gravity from the mountain causes the period to decrease.

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