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A tuning fork of frequency \(440 \mathrm{~Hz}\) is attached to a long string of linear mass density \(0 \cdot 01 \mathrm{~kg} \mathrm{~m}^{-1}\) kept under a tension of \(49 \mathrm{~N}\). The fork produces transverse waves of amplitude \(0-50 \mathrm{~mm}\) on the string. (a) Find the wave speed and the wavelength of the waves. (b) Find the maximum speed and acceleration of a particle of the string. (c) At what average rate is the tuning fork transmitting energy to the string?

Short Answer

Expert verified
a) Wave speed: 70 m/s, Wavelength: 0.1591 m. b) Maximum speed: 1.382 m/s, Maximum acceleration: 3.82 m/s². c) Power: 2.74 W.

Step by step solution

01

Finding the Wave Speed

To find the wave speed, we use the formula for wave speed on a string under tension:\[ v = \sqrt{\frac{T}{\mu}} \]where \( T = 49 \, \text{N} \) is the tension and \( \mu = 0.01 \, \text{kg/m} \) is the linear mass density.\[ v = \sqrt{\frac{49}{0.01}} = \sqrt{4900} = 70 \, \text{m/s} \]
02

Finding the Wavelength

The relationship between speed \( v \), frequency \( f \), and wavelength \( \lambda \) is given by:\[ v = f \lambda \]Given \( v = 70 \, \text{m/s} \) and \( f = 440 \, \text{Hz} \), solve for \( \lambda \):\[ \lambda = \frac{v}{f} = \frac{70}{440} \approx 0.1591 \, \text{m} \]
03

Finding Maximum Speed of a Particle on the String

The maximum speed \( v_{max} \) of a particle on the string is given by:\[ v_{max} = A \cdot \omega \]where \( A = 0.50 \times 10^{-3} \, \text{m} \) (converting from mm to m) is the amplitude, and \( \omega = 2 \pi f \).\[ \omega = 2 \pi \times 440 \approx 2764 \, \text{rad/s} \]\[ v_{max} = 0.50 \times 10^{-3} \times 2764 \approx 1.382 \, \text{m/s} \]
04

Finding Maximum Acceleration of a Particle on the String

The maximum acceleration \( a_{max} \) is given by:\[ a_{max} = A \cdot \omega^2 \]\[ a_{max} = 0.50 \times 10^{-3} \times (2764)^2 \approx 3.82 \times 10^3 \, \text{m/s}^2 \]
05

Finding Average Rate of Energy Transmission

The average power \( P \) transmitted by the tuning fork is:\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]Substitute the values:\[ \mu = 0.01, \ A = 0.50 \times 10^{-3}, \ \omega = 2764, \ v = 70 \]\[ P = \frac{1}{2} \times 0.01 \times (0.50 \times 10^{-3})^2 \times (2764)^2 \times 70 \approx 2.74 \, \text{W} \]

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Key Concepts

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

Wave Speed
Understanding wave speed is crucial in wave mechanics. The wave speed on a string depends on two primary factors: the tension in the string and the linear mass density. The formula used to calculate the wave speed \( v \) is given by \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension in the string and \( \mu \) is the linear mass density. These quantities look at how fast the wave travels along the string. In our exercise, we found the wave speed to be 70 m/s when using a tension of 49 N and a linear mass density of 0.01 kg/m. This demonstrates how a higher tension or lower mass density can lead to faster wave speeds.
Wavelength Calculation
To find the wavelength \( \lambda \) of a wave, we utilize the relationship between wave speed \( v \), frequency \( f \), and wavelength \( \lambda \): \[ v = f \lambda \]. This formula is very handy as it expresses how changes in frequency or wave speed affect the wavelength. Given that the wave speed is 70 m/s and the frequency of the tuning fork is 440 Hz, we can rearrange the formula to solve for the wavelength: \( \lambda = \frac{v}{f} \). Substituting the known values gives us a wavelength of approximately 0.159 meters. This outcome relies heavily on accurately knowing the wave's speed and frequency.
Energy Transmission
Energy transmission on a wave occurs through the medium with which it interacts, such as a string in our case. The rate at which energy is transmitted by the tuning fork to the string is determined using the average power \( P \). The formula for average power is given by: \[ P = \frac{1}{2} \mu A^2 \omega^2 v \], where \( \mu \) is the linear mass density, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( v \) is the wave speed. With the parameters from our problem, the average rate of energy transmission turns out to be about 2.74 watts. This calculation shows how energy efficiently transmits through oscillations on a string and is directly linked to the square of the wave's amplitude.
Linear Mass Density
Linear mass density \( \mu \) is a property of a one-dimensional object like a string, indicating how much mass is distributed per unit length. It's essential in determining wave properties such as wave speed. A lower mass density means less inertia, and hence the wave can travel faster. In our exercise, the linear mass density of the string was given as 0.01 kg/m. It's used directly in the calculation of wave speed and the power transmitted along the string. This fundamental property shows the importance of material characteristics in wave dynamics.
Particle Acceleration
When a wave travels through a medium, like a string, the particles of the medium oscillate. These oscillations include both speed and acceleration. The maximum acceleration \( a_{max} \) of a particle on the string is given by the formula: \[ a_{max} = A \omega^2 \], where \( A \) is the amplitude and \( \omega \) is the angular frequency. In this exercise, we found that the maximum acceleration is approximately 3820 m/s². This acceleration indicates how quickly a particle's velocity changes, reflecting the intensity and energy of the wave. It's pivotal to understand how acceleration affects wave motion and energy distribution.

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Most popular questions from this chapter

A scooter weighing \(150 \mathrm{~kg}\) together with its rider moving at \(36 \mathrm{~km} / \mathrm{hr}\) is to take a turn of radius \(30 \mathrm{~m}\). What horizontal force on the scooter is needed to make the turn possible?

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Two blocks each having a mass of \(3 \cdot 2 \mathrm{~kg}\) are connected by a wire \(C D\) and the system is suspended from the ceiling by another wire \(A B\) (figure 15 -E5). The linear mass density of the wire \(A B\) is \(10 \mathrm{~g} \mathrm{~m}^{-1}\) and that of \(\mathrm{CD}\) is \(8 \mathrm{~g} \mathrm{~m}^{-1}\). Find the speed of a transverse wave pulse produced in \(A B\) and in \(C D\).

The equation of a standing wave, produced on a string fixed at both ends, is $$ y=(0 \cdot 4 \mathrm{~cm}) \sin \left[\left(0-314 \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(600 \pi \mathrm{s}^{-1}\right) t\right] $$ What could be the smallest length of the string?

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