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Chapter 15: Problem 16

With what tension must a rope with length 2.50 \(\mathrm{m}\) and mass 0.120 \(\mathrm{kg}\) be stretched for transverse waves of frequency 40.0 \(\mathrm{Hz}\) to have a wavelength of 0.750 \(\mathrm{m} ?\)

Short Answer

Expert verified
43.2 N

Step by step solution

01

Calculate the wave speed

To calculate the wave speed, use the relation between wave speed, frequency, and wavelength given by: \( v = f \cdot \lambda \). Here, \( f = 40.0 \ \text{Hz} \) and \( \lambda = 0.750 \ \text{m} \). So, \( v = 40.0 \times 0.750 = 30.0 \ \text{m/s} \).
02

Calculate the linear mass density of the rope

The linear mass density \( \mu \) is the mass per unit length of the rope and is given by: \( \mu = \frac{m}{L} \). Here, \( m = 0.120 \ \text{kg} \) and \( L = 2.50 \ \text{m} \) , so \( \mu = \frac{0.120}{2.50} = 0.048 \ \text{kg/m} \).
03

Use wave speed to find the tension

The wave speed \( v \) on a stretched string is given by the equation \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the rope. Rearrange this to solve for \( T \): \( T = \mu v^2 \). Using \( \mu = 0.048 \ \text{kg/m} \) and \( v = 30.0 \ \text{m/s} \), we find \( T = 0.048 \times (30.0)^2 = 43.2 \ \text{N} \).

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Key Concepts

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

Wave Speed
Wave speed is an important concept when dealing with waves on strings or ropes. It indicates how fast a wave travels along the medium. The wave speed \( v \) is determined by the product of its frequency \( f \) and wavelength \( \lambda \) given by the equation:
  • \( v = f \times \lambda \)
For example, if you know the frequency of a wave is 40 Hz (hertz) and its wavelength is 0.750 m (meters), you can easily find the wave speed:
  • Calculate \( v = 40 \times 0.750 = 30 \) m/s (meters per second).
This tells you that the wave moves at a speed of 30 m/s through the rope. Understanding wave speed is crucial for determining other characteristics of waves like tension and linear mass density.
Linear Mass Density
Linear mass density, denoted by the symbol \( \mu \), is the mass per unit length of a rope or string. It plays a key role in determining the wave speed on the string, along with the tension applied to it. You can calculate linear mass density using the formula:
  • \( \mu = \frac{m}{L} \)
Where \( m \) is the mass of the rope, and \( L \) is its length. For instance, if a rope has a mass of 0.120 kg and a length of 2.50 m, the linear mass density will be:
  • \( \mu = \frac{0.120}{2.50} = 0.048 \) kg/m.
This value is the mass distributed over each meter of the rope, which helps in calculating the tension needed for wave propagation.
Transverse Waves
Transverse waves are a type of wave where the oscillations or particle movements occur perpendicular to the direction of the wave's travel. This is commonly seen in waves on a stretched rope or string.
  • These waves consist of peaks and troughs.
  • The speed at which these waves move is influenced by both the tension in the rope and its linear mass density.
To support transverse waves, a rope must be under tension. You can determine the tension using the wave speed equation:
  • \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, and \( \mu \) is the linear mass density.
Rearranging gives \( T = \mu v^2 \). For a wave speed of 30 m/s, if \( \mu = 0.048 \) kg/m, then tension \( T \) becomes:
  • \( T = 0.048 \times 30^2 = 43.2 \) N (Newtons).
This calculation shows that to produce and sustain transverse waves in the rope, a specific amount of tension is essential.

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Most popular questions from this chapter

Guitar String. One of the 63.5 -cm-long strings of an ordinary guitar is tuned to produce the note \(B_{3}(\) freguency 245 \(\mathrm{Hz})\) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is \(344 \mathrm{m} / \mathrm{s},\) find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

15.40\(\cdot\) A 1.50 -m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 48.0 \(\mathrm{m} / \mathrm{s}\) . What are the wavelength and frequency of (a) the fundamental: (b) the second overtone; (c) the fourth harmonic?

\mathrm{A} jet plane at takeoff can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2} at 30.0 \)\mathrm{m} away. But you prefer the tranquil 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 takeoff?

Combining Standing Waves. A guitar string of length \(L\) is plucked in such a way that the total wave produced is the sum of the fundamental and the second harmonic. That is, the standing wave is given by $$y(x, t)=y_{1}(x, t)+y_{2}(x, t)$$ where $$\begin{aligned} y_{1}(x, t) &=C \sin \omega_{1} t \sin k_{1} x \\ y_{2}(x, t) &=C \sin \omega_{2} t \sin k_{2} x \end{aligned}$$ with \(\omega_{1}=v k_{1}\) and \(\omega_{2}=v k_{2}\) . (a) At what values of \(x\) are the nodes of \(y_{1} ?\left(\) b) At what values of \(x\) are the nodes of \(y_{2} ?(\mathrm{c})\) Graph \right the total wave at \(t=0, t=\frac{1}{8} f_{1}, t=\frac{1}{4} f_{1}, t=\frac{3}{8} f_{1},\) and \(t=\frac{1}{2} f_{1}\) . (d) Does the sum of the two standing waves \(y_{1}\) and \(y_{2}\) produce a standing wave? Explain.

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g} .\) The wave speed is \(30.0 \mathrm{m} / \mathrm{s},\) and the wavelength is 0.200 \(\mathrm{m} .\) (a) If the wave is to have an average power of \(50.0 \mathrm{W},\) what must be the amplitude of the wave? (b) For this same string, if the amplitude and wave-length are the same as in part (a), what is the average power for the wave if the tension is increased such the wave speed is doubled?
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