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A \(6.00-\mathrm{kg}\) object oscillates back and forth at the end of a spring whose spring constant is 76.0 \(\mathrm{N} / \mathrm{m}\) . An observer is traveling at a speed of \(1.90 \times 10^{8} \mathrm{m} / \mathrm{s}\) relative to the fixed end of the spring. What does this observer measure for the period of oscillation?

Short Answer

Expert verified
The observed period of oscillation is approximately 2.272 s.

Step by step solution

01

Understand the Problem

We need to calculate the observed period of oscillation for a spring-mass system when the observer is moving at a relativistic speed. The known parameters are mass \(m = 6.00\,\mathrm{kg}\), spring constant \(k = 76.0\,\mathrm{N/m}\), and observer speed \(v = 1.90 \times 10^8 \,\mathrm{m/s}\).
02

Calculate the Period at Rest

The period \(T_0\) of an oscillating spring at rest is given by \(T_0 = 2\pi \sqrt{\frac{m}{k}}\). Substituting the given values:\[m = 6.00\,\mathrm{kg}, \quad k = 76.0\,\mathrm{N/m}\]\[T_0 = 2\pi \sqrt{\frac{6.00}{76.0}} \approx 1.759\,\mathrm{s}\]
03

Account for Relativistic Effects

As the observer is moving at a relativistic speed, we need to use time dilation, where the observed period \(T\) is given by:\[T = \frac{T_0}{\sqrt{1-\frac{v^2}{c^2}}}\]Here, \(c = 3.00 \times 10^8 \,\mathrm{m/s}\) is the speed of light. Substituting the known values:\[T = \frac{1.759}{\sqrt{1 - \frac{(1.90 \times 10^8)^2}{(3.00 \times 10^8)^2}}}\]
04

Perform the Calculation

First, calculate \(v^2/c^2\):\[\frac{(1.90 \times 10^8)^2}{(3.00 \times 10^8)^2} = 0.401decrease-\]Now, calculate the square root term:\[\sqrt{1 - 0.401} = \sqrt{0.599} \approx 0.774\]The observed period is:\[T = \frac{1.759}{0.774} \approx 2.272\,\mathrm{s}\]
05

State the Final Result

The observed period of oscillation for the moving observer is approximately \(2.272\,\mathrm{s}\).

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Key Concepts

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

Harmonic Oscillation
Harmonic oscillation refers to the repeating back and forth motion observed in various physical systems, such as pendulums or springs. This motion can typically be described by simple mathematical relations involving sine or cosine functions. In the context of a spring-mass system, an object attached to a spring undergoes harmonic motion as it moves back and forth from its equilibrium position.

An important property of harmonic oscillation is its periodic nature, meaning the motion repeats itself at regular intervals, known as the period. This is predominantly defined by the mass attached and the spring constant, a measure of the spring's stiffness. The period is independent of the amplitude for small oscillations, which means the size of the oscillation doesn't change the length of each complete cycle.

For the spring-mass system in our exercise, we calculate the period at rest using the formula:
  • \( T_0 = 2\pi \sqrt{\frac{m}{k}} \)
  • Where \( m \) is the mass and \( k \) is the spring constant.
    For a mass of 6.00 kg and a spring constant of 76.0 N/m, this gives a period \( T_0 \approx 1.759 \, \mathrm{s} \).
Spring-Mass System
The spring-mass system is a classic model in physics that helps us understand harmonic motion. It consists of a mass attached to a spring, where the mass can oscillate back and forth as the spring compresses and extends. This model is governed by Hooke's Law, which states that the force exerted by the spring is proportional to its deformation, and is expressed as:
  • \( F = -kx \)
  • Where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
When displaced from the equilibrium, the spring force works to restore the object to its original position, resulting in oscillatory motion. The study of this system involves understanding key concepts like kinetic and potential energy, as the energy transforms continuously between these two forms during oscillation.

The natural frequency of the system, which is dependent on the mass and spring constant, dictates how fast the system oscillates in the absence of external forces or damping. For different system setups, spring-mass systems are invaluable for exploring both fundamental mechanical properties and more advanced concepts, such as resonance and damping, in complex oscillatory systems.
Time Dilation in Relativity
Time dilation is a fascinating concept from Einstein's theory of relativity. It describes how time experienced by an observer in motion differs from that experienced by a stationary observer. When dealing with high speeds, especially those approaching the speed of light, time dilation becomes noticeable. This effect means that from the perspective of a moving observer, time appears to pass slower.

In our problem, the observer is traveling at a significant fraction of the speed of light, specifically at \( v = 1.90 \times 10^8 \, \mathrm{m/s} \). To determine how this affects the period of the spring-mass system, we use the time dilation formula:
  • \( T = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}} \)
  • Here, \( T_0 \) is the period at rest, and \( c \) is the speed of light \( 3.00 \times 10^8 \, \mathrm{m/s} \).
Applying these values, the moving observer sees the period increase to approximately 2.272 seconds, showcasing the impact of relativistic effects. This fundamental concept not only underscores the elasticity of time in physics but also highlights the interconnectedness of energy, motion, and spacetime.

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Most popular questions from this chapter

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