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Chapter 16: Problem 61

In an experiment on standing waves, a string \(90 \mathrm{~cm}\) long is attached to the prong of an electrically driven tuning fork that oscillates perpendicular to the length of the string at a frequency of \(60 \mathrm{~Hz}\). The mass of the string is \(0.044 \mathrm{~kg}\). What tension must the string be under (weights are attached to the other end) if it is to oscillate in four loops?

Short Answer

Expert verified
The tension required is approximately 35.53 N.

Step by step solution

01

Understand the Physical Setup

The string is oscillating with standing waves in four loops or segments. This means that there are four antinodes along the string. You are given the string length \( L = 90 \text{ cm} = 0.9 \text{ m} \), frequency \( f = 60 \text{ Hz} \), and mass \( m = 0.044 \text{ kg} \).
02

Determine the Wavelength

For standing waves, the wavelength \( \lambda \) is related to the length of the string and the number of loops \( n \). Since there are 4 loops, there are 4 half-wavelengths along the string:\[ \frac{n\lambda}{2} = L \]\[ \frac{4\lambda}{2} = 0.9 \text{ m} \]\[ 2\lambda = 0.9 \text{ m} \]Solving for \( \lambda \), we find: \( \lambda = 0.45 \text{ m} \).
03

Calculate the Wave Speed on the String

The wave speed \( v \) on the string is determined by the relationship:\[ v = f \lambda \]\[ v = 60 \times 0.45 \text{ m/s} \]\[ v = 27 \text{ m/s} \].
04

Find the Linear Density of the String

The linear density \( \mu \) of the string is given by its mass divided by its length:\[ \mu = \frac{m}{L} \]\[ \mu = \frac{0.044}{0.9} \text{ kg/m} \]\[ \mu \approx 0.0489 \text{ kg/m} \].
05

Determine the Tension Needed

The tension \( T \) in the string is related to the wave speed \( v \) and linear density \( \mu \) by:\[ v = \sqrt{\frac{T}{\mu}} \]Rearranging to solve for \( T \):\[ T = v^2 \mu \]\[ T = (27)^2 \times 0.0489 \]\[ T = 35.5321 \text{ N} \].
06

Verify and Conclude

Re-examine each calculation to ensure the proper conversion and application of formulas. The tension \( T \) that provides four loops or segments along a 90 cm string at a driving frequency of 60 Hz is \( 35.53 \text{ N} \), rounded to two decimal places.

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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 studying waves on a string, understanding tension is crucial. Tension is essentially the force applied along the string to keep it taut. It plays a significant role in wave formation.
In the context of standing waves, this tension is responsible for the way waves propagate across the string. The tension affects the speed and shape of the waves, determining the frequency and mode of oscillation.
To calculate the tension needed to create specific wave patterns or modes鈥攕uch as the four loops in this exercise鈥攚e use the formula:
  • \[ T = v^2 \mu \]
This formula links tension \( T \) to wave speed \( v \) and linear density \( \mu \) of the string, providing a mathematical understanding of the relationship between these elements.
In our exercise, calculating this tension allows for precise adjustment of the force required to achieve the desired wave pattern.
Wave Speed
Wave speed is the rate at which waves travel through a medium鈥攊n this case, a string. It is crucial in defining how quickly the wave patterns form along the string.
For a string attached to a tuning fork, wave speed depends on both the frequency of the fork and the wavelength of the waves produced. The wave speed can be calculated using the formula:
  • \[ v = f \lambda \]
Where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength.
In our example, with a frequency of 60 Hz and a wavelength of 0.45 m, the wave speed is found to be 27 m/s. This speed is crucial for calculating other forces, such as tension, that influence the waves.
Linear Density
Linear density is a measure of mass distribution along the length of a string. It is defined as mass per unit length and can significantly impact how waves behave.
The formula for linear density \( \mu \) is:
  • \[ \mu = \frac{m}{L} \]
Where \( m \) is the mass of the string, and \( L \) is its length.
Understanding linear density is important because it directly affects calculations of wave speed and tension. For instance, a heavier or denser string may require more tension to achieve the same wave speed as a lighter one.
In our example, the string has a mass of 0.044 kg and is 0.9 m long, leading to a linear density of approximately 0.0489 kg/m. This value is a key component in determining the tension needed for standing waves and is instrumental in configuring the wave speed on the string.

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Most popular questions from this chapter

A string under tension \(\tau_{i}\) oscillates in the third harmonic at frequency \(f_{3}\), and the waves on the string have wavelength \(\lambda_{3}\). If the tension is increased to \(\tau_{f}=4 \tau_{i}\) and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of \(f_{3}\) and \((b)\) the wavelength of the waves in terms of \(\lambda_{3} ?\)

\(A 1.50 \mathrm{~m}\) wire has a mass of \(8.70 \mathrm{~g}\) and is under a tension of \(120 \mathrm{~N}\). The wire is held rigidly at both ends and set into oscillation. (a) What is the speed of waves on the wire? What is the wavelength of the waves that produce (b) one-loop and (c) twoloop standing waves? What is the frequency of the waves that produce (d) one-loop and (e) two-loop standing waves?

Consider a loop in the standing wave created by two waves (amplitude \(5.00 \mathrm{~mm}\) and frequency \(120 \mathrm{~Hz}\) ) traveling in opposite directions along a string with length \(2.25 \mathrm{~m}\) and mass \(125 \mathrm{~g}\) and under tension \(40 \mathrm{~N}\). At what rate does energy enter the loop from (a) each side and (b) both sides? (c) What is the maximum kinetic energy of the string in the loop during its oscillation?

The tension in a wire clamped at both ends is doubled without appreciably changing the wire's length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire?

Two identical traveling waves, moving in the same direction, are out of phase by \(\pi / 2\) rad. What is the amplitude of the resultant wave in terms of the common amplitude \(y_{m}\) of the two combining waves?
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