/*! 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 51 How fast (as a percentage of lig... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 51

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we recelve from it is 10.0\(\%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

Short Answer

Expert verified
The star moves at about 9.5% of light speed towards us.

Step by step solution

01

Understand Doppler Effect for Light

The Doppler Effect for light relates the observed frequency \( f' \) to the emitted frequency \( f \) using the equation: \( \frac{f'}{f} = \sqrt{\frac{1+\beta}{1-\beta}} \), where \( \beta = \frac{v}{c} \), and \( c \) is the speed of light.
02

Set Up the Equation

We know that \( f' = 1.10f \) since the frequency is 10\(\%\) higher. Therefore, \( \frac{f'}{f} = 1.10 \), and we set this equal to the right side of the Doppler Effect formula: \( 1.10 = \sqrt{\frac{1+\beta}{1-\beta}} \).
03

Square Both Sides

Square both sides of the equation to eliminate the square root: \( 1.10^2 = \frac{1+\beta}{1-\beta} \). This simplifies to \( 1.21 = \frac{1+\beta}{1-\beta} \).
04

Solve for \( \beta \)

Cross multiply to solve for \( \beta \): \( 1.21(1-\beta) = 1+\beta \). This leads to \( 1.21 - 1.21 \beta = 1 + \beta \). Rearranging gives \( 0.21 = 2.21 \beta \). Solve for \( \beta \) by dividing: \( \beta = \frac{0.21}{2.21} \approx 0.095 \).
05

Determine Direction of Motion

Since the frequency observed is higher than emitted, the star is moving towards us, as approaching objects increase frequency (blueshift).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Light Frequency
Light frequency refers to how often the waves of light pass a certain point per second, measured in Hertz (Hz). This frequency determines the color of visible light and plays a crucial role in understanding phenomena such as the Doppler Effect.

Within the context of the Doppler Effect, when we observe a star whose light frequency is altered, it suggests that the star is moving relative to us. If the observed frequency is higher than what is typically emitted, it indicates a "blueshift." Conversely, a lower frequency would indicate a "redshift," suggesting that the object is moving away.

Adjustments in light frequency help astronomers determine the motion of stars and galaxies relative to Earth. Here, the challenge is predicting the star's velocity when observing a specific frequency increase like 10\( \% \). This gives insight into the star's speed and direction in space.
Relative Velocity
Relative velocity is the speed of an object in relation to another object. In this case, we are interested in the velocity of a star concerning Earth. This concept is crucial in understanding the Doppler Effect for light, as it's the relative motion that causes shifts in light frequency.

In the given exercise, we determine how fast a star needs to travel for its light frequency to increase by 10\( \% \). By using the Doppler Effect formula, we gauge this speed as a portion of the speed of light. This precise calculation allows physicists to measure cosmic movements over vast distances, where direct measurement is impossible.
  • A velocity toward the observer leads to a frequency increase (blueshift).
  • A velocity away from the observer causes a frequency decrease (redshift).
Blueshift
Blueshift occurs when the light from an object in space, like a star, has a higher frequency when observed compared to when it was emitted. This happens when the object is moving towards the observer, compressing the light waves and making them shift towards the blue end of the visible spectrum.

In the exercise, a 10\( \% \) increase in light frequency means the perceived light is blueshifted since the star is approaching Earth. The concept of blueshift is a direct application of the Doppler Effect, illustrating the relationship between the motion of celestial bodies and the frequency of light.
  • Blueshift indicates movement toward the Earth.
  • It allows astronomers to infer the drift and speed of stars and galaxies.
Speed of Light
The speed of light, denoted as \( c \), is a fundamental constant of nature, approximately equal to 299,792,458 meters per second (or about 300,000 kilometers per second). In physics, it is of paramount importance, especially in the realm of relativity and electromagnetic theory.

When calculating relative velocity using the Doppler Effect, the speed of light serves as a reference. As seen in the exercise, the star's speed is expressed as a percentage of this constant.

This choice of unit is incredibly helpful because it lets scientists express velocities on a universal scale. By understanding light speed, we can also comprehend cosmic distances and the expansive universe.
  • The speed of light is a crucial factor in the Doppler Effect calculations.
  • It functions as the ultimate speed limit in the universe.
  • Describing speeds as a percentage of the speed of light helps compare different motions in space effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A soprano and a bass are singing a duet. While the soprano \(\operatorname{sings}\) an \(A^{\prime \prime}\) at 932 . Hz, the bass sings an \(A^{\prime \prime}\) but three octaves lower. In this concert hall, the density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its bulk modulus is \(1.42 \times 10^{5} \mathrm{Pa}\) . In order for their notes to have the same Rnund intensity level, what must he (a) the ratio of the pressur amplitude of the bass to that of the soprano, and (b) the ratio of the displacement amplitude of the bass to that of the soprano? (c) What displacement amplitude (in \(\mathrm{m}\) and \(\mathrm{nm}\) ) does the soprano produce to sing her \(A^{\prime \prime}\) at 72.0 \(\mathrm{dB} ?\)

A swimming duck paddles the water with its feet once cvery 1.6 \(\mathrm{s}\) , producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 \(\mathrm{m} / \mathrm{s}\) , and the crests of the waves ahead of the duck are spaced 0.12 \(\mathrm{m}\) apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?

Two loudspeakers, \(A\) and \(B,\) radiate sound uniformly in all directions in air at \(20^{\circ} \mathrm{C}\) . The acoustic power output from \(A\) is \(8.00 \times 10^{-4} \mathrm{W}\) , and from \(B\) it is \(6.00 \times 10^{-5} \mathrm{W}\) . Both loudspeakcrs are vibrating in phase at a frequency of 172 \(\mathrm{Hz}\) (a) Determine the difference in phase of the two signals at a point \(C\) along the line joining \(A\) and \(B, 3.00 \mathrm{m}\) from \(B\) and 4.00 \(\mathrm{m}\) from \(A(\mathrm{Fig} .16 .46)\) . (b) Determine the intensity and sound intensity level at \(C\) from speaker \(A\) if speaker \(B\) is turned off and the intensity and sound intensity level at point \(C\) from speaker \(B\) if speaker \(A\) is turned off. (c) With both speakers on, what are the intensity and sound intensity level at \(C ?\)

An organ pipe has two successive harmonics with frequencies 1372 and 1764 \(\mathrm{Hz}\) (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 \(\mathrm{m} / \mathrm{s}\) when the gas temperature is \(22.0^{\circ} \mathrm{C}\) . For a certain experiment, you need to have the same uscillator produce sound of waveleugth 28.5 \(\mathrm{cm}\) in this gas. What should the gas temperature be to achieve this wavelength?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Force

Read Explanation

Oscillations

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.