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Chapter 16: Problem 26

A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(260 \mathrm{~g}\). The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) for the average power to be \(85.0 \mathrm{~W} ?\)

Short Answer

Expert verified
The frequency must be approximately 159 Hz.

Step by step solution

01

Calculate Linear Density

First, calculate the linear density \( \mu \) of the string, which is the mass per unit length. Given that the mass \( m \) of the string is \( 260 \text{ g} = 0.260 \text{ kg} \) and the length \( L \) is \( 2.70 \text{ m} \), the linear density is:\[ \mu = \frac{m}{L} = \frac{0.260 \text{ kg}}{2.70 \text{ m}} = 0.0963 \text{ kg/m} \]
02

Determine Wave Speed

Next, calculate the speed of waves \( v \) on the string using the formula:\[ v = \sqrt{\frac{T}{\mu}} \]where \( T \) is the tension in the string \( 36.0 \text{ N} \). Substituting the values, we get:\[ v = \sqrt{\frac{36.0 \text{ N}}{0.0963 \text{ kg/m}}} = \sqrt{373.88} \approx 19.34 \text{ m/s} \]
03

Relate Power with Amplitude and Frequency

The average power \( P \) carried by the waves is related to the frequency \( f \), amplitude \( A \), wave speed \( v \), and linear density \( \mu \) by the formula:\[ P = \frac{1}{2} \mu v \omega^2 A^2 \]where \( \omega \) is the angular frequency given by \( \omega = 2\pi f \). Since \( A = 7.70 \text{ mm} = 0.00770 \text{ m} \), substitute the known values:\[ 85.0 \text{ W} = \frac{1}{2} \times 0.0963 \text{ kg/m} \times 19.34 \text{ m/s} \times (2\pi f)^2 \times (0.00770 \text{ m})^2 \]
04

Solve for Frequency

Rearrange the power equation to solve for \( f \):\[ 85.0 = \frac{1}{2} \times 0.0963 \times 19.34 \times 4\pi^2 f^2 \times 0.00770^2 \]This simplifies to:\[ f^2 = \frac{85.0}{\frac{1}{2} \times 0.0963 \times 19.34 \times 4\pi^2 \times 0.00770^2} \]Calculate the frequency:\[ f^2 \approx 25317.6 \Rightarrow f \approx \sqrt{25317.6} \approx 159 \text{ Hz} \]

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Key Concepts

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

Tension in a String
When we talk about tension in a string, we're referring to the force that's applied along the length of the string. This force keeps the string taut, allowing waves to travel through it.
The tension in the string is crucial because it directly affects the speed at which waves travel. Think of tension like tightening a guitar string; the tighter it is, the higher the pitch of sound it makes when plucked.
For a string with tension, the force is measured in newtons (N). In the exercise, the given tension is 36.0 N. This tension acts to keep the string straight and enables it to carry waves efficiently.
  • Waves on a string are influenced by how tightly the string is pulled (tension).
  • A string under higher tension will have waves that travel faster.
Keeping these ideas in mind helps us understand how tension plays a role in wave mechanics.
Linear Density of a String
Linear density is an important concept when dealing with waves on a string. It is essentially the mass of the string per unit length, measured in kilograms per meter (kg/m).
To find linear density (\(\mu\)), divide the mass of the string by its total length.
In the exercise, the string has a mass of 260 grams (or 0.260 kg when converted to kilograms), and its length is 2.70 meters. Thus, the linear density is calculated as:\[\mu = \frac{0.260 \text{ kg}}{2.70 \text{ m}} = 0.0963 \text{ kg/m}\]This value helps us understand how the mass of the string is distributed along its length.
  • Linear density affects how waves travel on a string.
  • A string with lower linear density will have faster traveling waves if tension is constant.
Understanding linear density is crucial for calculations involving wave speed and energy transmission.
Wave Speed Calculation
Wave speed calculation allows us to determine how fast a wave travels along a string.
This speed depends on both the tension in the string and its linear density, and can be calculated using the formula:\[ v = \sqrt{\frac{T}{\mu}} \]where \(v\) is the wave speed, \(T\) is the tension, and \(\mu\) is the linear density. Using the given values from the exercise:\[ v = \sqrt{\frac{36.0 \text{ N}}{0.0963 \text{ kg/m}}} \approx 19.34 \text{ m/s} \]This equation shows us that wave speed increases with higher tension and decreases with higher linear density. Therefore, both the tension and linear density combine to influence how quickly a wave can travel on a string.
  • Wave speed is influenced by tension and linear density based on a square root relationship.
  • Increasing tension increases wave speed, while increasing linear density reduces speed.
Learning how to calculate wave speed is fundamental for understanding wave dynamics on strings.

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Most popular questions from this chapter

Two sinusoidal waves of the same wavelength travel in the same direction along a stretched string. For wave \(1, y_{m}=3.0 \mathrm{~mm}\) and \(\phi=\) \(0 ;\) for wave \(2, y_{m}=5.0 \mathrm{~mm}\) and \(\phi=70^{\circ} .\) What are the (a) amplitude and (b) phase constant of the resultant wave?

A wave on a string is described by $$ y(x, t)=15.0 \sin (\pi x / 8-4 \pi t), $$ where \(x\) and \(y\) are in centimeters and \(t\) is in seconds. (a) What is the transverse speed for a point on the string at \(x=6.00 \mathrm{~cm}\) when \(t=0.250 \mathrm{~s}\) ? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a point on the string at \(x=6.00 \mathrm{~cm}\) when \(t=0.250 \mathrm{~s}\) ? (d) What is the magnitude of the maximum transverse acceleration for any point on the string?

Two sinusoidal waves of the same frequency are to be sent in the same direction along a taut string. One wave has an amplitude of \(5.0 \mathrm{~mm}\), the other \(8.0 \mathrm{~mm}\). (a) What phase difference \(\phi_{1}\) between the two waves results in the smallest amplitude of the resultant wave? (b) What is that smallest amplitude? (c) What phase difference \(\phi_{2}\) results in the largest amplitude of the resultant wave? (d) What is that largest amplitude? (e) What is the resultant amplitude if the phase angle is \(\left(\phi_{1}-\phi_{2}\right) / 2 ?\)

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a wire that is \(10.0 \mathrm{~m}\) long, has a mass of \(100 \mathrm{~g}\), and is stretched under a tension of \(250 \mathrm{~N}\) ?

A sinusoidal transverse wave traveling in the negative direction of an \(x\) axis has an amplitude of \(1.00 \mathrm{~cm}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m},(\) b) \(\omega,(\mathrm{c}) k\), and \((\mathrm{d})\) the correct choice of sign in front of \(\omega ?\)
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