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Chapter 15: Problem 16

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

Short Answer

Expert verified
The tension must be 43.20 N.

Step by step solution

01

Understand the Wave Equation

Transverse wave speed \( v \) is determined by the product of the wave's frequency \( f \) and its wavelength \( \lambda \). The equation is: \( v = f \times \lambda \). We will use this equation to find the wave speed.
02

Calculate the Wave Speed

Use the given frequency \( f = 40.0 \) Hz and wavelength \( \lambda = 0.750 \) m to calculate the wave speed \( v \). Substitute these values into the wave equation: \( v = 40.0 \times 0.750 \). Thus, \( v = 30.0 \) m/s.
03

Use the Wave Speed in the Tension Formula

The speed of a transverse wave on a string is given by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the linear mass density of the string. We'll solve this equation to find the tension \( T \).
04

Calculate the Linear Mass Density

The linear mass density \( \mu \) is the mass of the string divided by its length. Calculate \( \mu \) using the formula: \( \mu = \frac{m}{L} = \frac{0.120}{2.50} = 0.048 \) kg/m.
05

Solve for Tension

Rearrange the wave speed formula \( v = \sqrt{\frac{T}{\mu}} \) to solve for \( T \), giving \( T = v^2 \times \mu \). Substitute \( v = 30.0 \) m/s and \( \mu = 0.048 \) kg/m into the equation: \( T = 30.0^2 \times 0.048 = 43.20 \) N.

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the motion of the medium is perpendicular to the direction of the wave. These waves can be visualized as ripples on the surface of water or the oscillations of a plucked string. Unlike longitudinal waves, which move parallel to the wave's direction (like sound waves through the air), transverse waves showcase side-to-side or up-and-down movement.

Understanding transverse waves is key to grasping wave dynamics in strings or any medium supporting this motion. In transverse waves, the energy is transferred through oscillations perpendicular to the wave direction, making them unique and easily recognizable in various physical phenomena.
Wave Equation
The wave equation is fundamental in determining many properties of waves. It can be expressed as: \[ v = f \times \lambda \] where
  • \(v\) is the wave speed,
  • \(f\) is the frequency, and
  • \(\lambda\) is the wavelength.
This simple yet powerful equation shows that the speed of a wave is the product of its frequency and wavelength.

In practical applications, knowing two of these components allows us to calculate the third. For instance, if you know the frequency and wavelength, you can easily find how fast the wave is traveling. This relationship is crucial when dealing with any wave-related phenomena, such as light, sound, or vibrations on a string.
Linear Mass Density
Linear mass density, denoted by \( \mu \), is a measure of the mass per unit length of a material, specifically in a string context. The formula to calculate it is: \[ \mu = \frac{m}{L}\] where
  • \(m\) is the total mass, and
  • \(L\) is the length of the string.
This parameter is essential when exploring wave behavior in strings, as it directly affects the wave speed and tension that can be maintained.

For instance, in our example, with a mass of 0.120 kg and a length of 2.50 m, the linear mass density is calculated as 0.048 kg/m. A higher linear mass density typically means the wave will travel slower, given the same tension, because there's more mass that the wave has to move along the medium.
Tension Calculation
Calculating tension is a critical step in determining how tightly a string must be stretched to support a specific wave speed. The formula that connects wave speed with tension is: \[ v = \sqrt{\frac{T}{\mu}} \] where
  • \(v\) is the wave speed,
  • \(T\) is the tension in the string, and
  • \(\mu\) is the linear mass density.
To find the tension \(T\), this formula can be rearranged to: \[ T = v^2 \times \mu \]
This shows that tension depends on both the wave speed and the linear mass density. In our example, using a wave speed of 30.0 m/s and a linear mass density of 0.048 kg/m, the calculated tension is 43.20 N. Ensuring accurate tension is vital in applications such as musical instruments or engineering implementations, where precise wave propagation is necessary.

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Most popular questions from this chapter

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t) = (5.60 \, \mathrm{cm}) \mathrm{sin} [(0.0340 \mathrm{rad/cm})x] \mathrm{sin} [(150.0 \, \mathrm{rad/s})t]\), where the origin is at the left end of the string, the x-axis is along the string, and the \(y\)-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x,t)\) for this string if it were vibrating in its eighth harmonic?

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda /4\), and (iii) \(x = \lambda /8\), from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

A jet plane at takeoff can produce sound of intensity 10.0 W/m\(^2\) at 30.0 m away. But you prefer the tranquil sound of normal conversation, which is 1.0 \(\mu\)W/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?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

For a string stretched between two supports, two successive standing-wave frequencies are 525 Hz and 630 Hz. There are other standing-wave frequencies lower than 525 Hz and higher than 630 Hz. If the speed of transverse waves on the string is 384 m/s, what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.
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