/*! 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 20 A piano wire with mass 3.00 \(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

Short Answer

Expert verified
The average power is approximately 0.534 W. If the amplitude is halved, the average power is reduced to 0.1335 W.

Step by step solution

01

Convert Units

Convert the mass from grams to kilograms and the wire length from centimeters to meters. The mass is 3.00 g which is equal to 0.003 kg, and the length is 80.0 cm which is 0.8 m.
02

Calculate Linear Density

The linear density \( \mu \) of the wire is mass per unit length. Calculate it using the formula:\[ \mu = \frac{m}{L} = \frac{0.003}{0.8} = 0.00375 \, \mathrm{kg/m} \]
03

Determine Wave Speed

The wave speed \( v \) on the wire can be calculated using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T = 25.0 \, \mathrm{N} \). Substitute the values to find \( v \):\[ v = \sqrt{\frac{25.0}{0.00375}} \approx 81.64 \, \mathrm{m/s} \]
04

Calculate Wave Number and Angular Frequency

The wave number \( k \) and angular frequency \( \omega \) are given by \( k = \frac{2\pi}{\lambda} \) and \( \omega = 2\pi f \). Using the frequency \( f = 120.0 \, \mathrm{Hz} \), calculate \( \omega \):\[ \omega = 2\pi \times 120 = 240\pi \, \mathrm{rad/s} \]
05

Calculate Average Power with Original Amplitude

The average power \( P \) carried by a wave is given by \( P = \frac{1}{2} \mu \omega^2 A^2 v \). Using \( A = 1.6 \, \mathrm{mm} = 0.0016 \, \mathrm{m} \), substitute the values to calculate the power:\[ P = \frac{1}{2} \times 0.00375 \times (240\pi)^2 \times (0.0016)^2 \times 81.64 \approx 0.534 \, \mathrm{W} \]
06

Effect of Halving the Amplitude

If the amplitude \( A \) is halved, the new amplitude \( A' \) is 0.0008 m. Substitute \( A' \) into the power formula: \[ P' = \frac{1}{2} \mu \omega^2 (A')^2 v = \frac{1}{2} \mu \omega^2 \left(\frac{A}{2}\right)^2 v = \frac{1}{4} \times 0.534 = 0.1335 \, \mathrm{W} \].

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 91Ó°ÊÓ!

Key Concepts

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

Wave Speed
Wave speed is a fundamental concept in physics, describing how fast a wave travels through a medium. For waves on a stretched string, like the piano wire in our problem, the wave speed can be determined using the tension in the string and its linear density. This relationship is expressed with the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in Newtons and \( \mu \) is the linear density in kilograms per meter.
In our specific example, the wave speed on the piano wire was calculated to be approximately 81.64 meters per second. Understanding how to manipulate this formula is crucial for solving problems regarding wave speed in various physical contexts. It shows how wave speed is directly influenced by changes in tension and linear density.
  • High tension leads to faster wave speed.
  • Greater linear density results in slower wave speed.
These insights help us understand wave motion in strings, cables, and even sound waves in air.
Linear Density
Linear density is a measure of how much mass is distributed along a unit length of string or wire. It is calculated using the formula \( \mu = \frac{m}{L} \), where \( m \) is the mass and \( L \) is the length. For the piano wire in the exercise, linear density is found to be 0.00375 kg/m after converting from grams and centimeters to kilograms and meters.
Linear density plays an important role in understanding wave behavior, especially for waves traveling along stretched strings or wires. It affects the wave speed and the tension needed to achieve a certain pitch. This is particularly relevant in musical instruments where precision in linear density can influence sound quality.
  • Lower linear density means less mass per unit length, potentially increasing wave speed.
  • Higher linear density, conversely, reduces wave speed, affecting how quickly vibration travels.
When solving problems involving linear density, always convert units correctly to ensure accurate calculations.
Angular Frequency
Angular frequency is a measure of how fast something oscillates. It's given by the formula \( \omega = 2\pi f \), where \( f \) is the frequency in Hertz. In our case, with a frequency of 120 Hz, the angular frequency is calculated as \( 240\pi \) rad/s.
Understanding angular frequency is crucial for any exercise involving periodic motion, like waves on strings or sound waves. It helps describe the wave's temporal characteristics, or how fast the wave cycle repeats per unit time. In the formula for calculating power, \( \omega \) reflects the rapidity of the wave oscillations:
  • Higher angular frequency means more oscillations in a given time, implying the wave transfers energy quicker.
  • Angular frequency is central in calculating other properties like wave energy and power.
Knowledge of angular frequency is essential when you deal with wave power and energy transmission in various media.
Wave Amplitude
Wave amplitude refers to the maximum displacement of points on a wave from its undisturbed position, often in meters. It indicates the energy transported by the wave: larger amplitudes mean more energy. In this problem, the initial amplitude was 1.6 mm (or 0.0016 m).
Amplitude is a crucial factor when considering wave energy and power. The wave's power is proportional to the square of the amplitude. Therefore, if you double the amplitude, the power increases fourfold. Conversely, halving the amplitude reduces the power by a factor of four, as shown in the exercise outcome when the amplitude was halved.
  • Higher amplitude waves carry more energy and power, impacting the system they interact with more forcefully.
  • Lower amplitude indicates less energy transfer, which might be useful in minimizing impact.
This knowledge is particularly important in sound and light waves, affecting what we hear and see.

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

Adjacent antinodes of a standing wave on a string are 15.0 \(\mathrm{cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 \(\mathrm{cm}\) and period 0.0750 \(\mathrm{s}\) . The string lies along the \(+x\) -axis and is fixed at \(x=0 .(\text { a) How far }\) apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

A guitar string is vibrating in its fundamental mode, with nodes at each end. The length of the segment of the string that is free to vibrate is 0.386 \(\mathrm{m}\) . The maximum transverse acceleration of a point at the middle of the segment is \(8.40 \times 10^{3} \mathrm{m} / \mathrm{s}^{2}\) and the maximum transverse velocity is 3.80 \(\mathrm{m} / \mathrm{s}\) (a) What is the amplitude of this standing wave? (b) What is the wave speed for the transverse traveling waves on this string?

A uniform rope with length \(L\) and mass \(m\) is held at one end and whirled in a horizontal circle with angular velocity \(\omega\) . You can ignore the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.

A piano tuner stretches a steel piano wire with a tension of 800 \(\mathrm{N}\) . The steel wire is 0.400 \(\mathrm{m}\) long and has a mass of 3.00 \(\mathrm{g}\) . (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{Hz} ?\)

A jet plane at take-off can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2}\) at 30.0 \(\mathrm{m}\) away. But you prefer the tranguil sound of normal conversation, which is 1.0\(\mu \mathrm{W} / \mathrm{m}^{2}\) . Assume that the plane behaves like a point source of sound. (a) What is the closest distance you should live from the airport runway to preserve your peace of mind? (b) What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? (c) What power of sound does the jet produce at take-off?

See all solutions

Recommended explanations on Physics Textbooks

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.