91Ó°ÊÓ

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 \(M g\) 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 standing-wave pattern and obtain the following data: $$ \begin{array}{l|lllll} \text { String } & \text { A } & \text { A } & \text { B } & \text { B } & \text { C } \\ \hline \mu(\mathrm{g} / \mathrm{cm}) & 0.0260 & 0.0260 & 0.0374 & 0.0374 & 0.0482 \\ M(\mathrm{~g}) & 559 & 249 & 365 & 207 & 262 \\ d(\mathrm{~cm}) & 48.1 & 31.9 & 32.0 & 24.2 & 23.8 \end{array} $$ (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^{2}(\) in \(\mathrm{kg} \cdot \mathrm{m})\) versus \(M(\) in \(\mathrm{kg}) .\) Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straightline fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). (d) For string A \((\mu=0.0260 \mathrm{~g} / \mathrm{cm}),\) what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of \(24.0 \mathrm{~cm}\) ? Use the value of \(f\) that you calculated in part (c).

Step by step solution

01

Understand the Problem

The first step involves understanding the given problem and the principles involved. We know that wave speed \(v\) is given by \(v = \sqrt{T/μ}\), where \(T\) is the tension in the string and \(μ\) is the string's linear mass density. Since we have a standing wave, the distance between nodes \(d\) is equal to one wavelength. In turn, we know that wave speed equals frequency multiplied by wavelength. We can therefore derive the equation \(d = v/f = \sqrt{T/μ} / f\). After squaring both sides, we get \(d² = T/(f²μ)\).

02

Graphical Analysis

We're instructed to graph \(\mu d²\) versus \(M\). According to the derived equation, \(\mu d²\) should equal \(T/f²\), and since the tension \(T\) is \(Mg\), we expect our graph to be a straight line with a slope equal to \(g/f²\). This means the data plotted in this way should fall close to a straight line.

03

Determine the Frequency

We use the slope of the best-fit straight line to determine the frequency \(f\) of the waves produced on the string by the oscillator. The slope should equal \(g/f²\), so we can rearrange the equation to solve for \(f\). Here, we get \(f = \sqrt{g/\text{{slope}}}\). Given the gravity \(g = 9.8 m/s²\), we need the slope from the graph.

04

Find the Required Mass

For string A (\(μ = 0.026 g/cm\)), we need to find the mass \(M\) required to produce a standing wave with a node-to-node distance of \(24.0 cm\). We use the formula \(d² = T/(f²μ)\) to solve for \(T\), then solve the equation \(T = Mg\) for \(M\), giving us \(M = d²f²μ/g\). Substituting given values and calculated \(f\) will yield the required mass \(M\).

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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 a key concept in understanding how waves travel along a string. When dealing with a string, the wave speed \( v \) can be calculated by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) represents the tension in the string, and \( \mu \) is the string's linear mass density. This equation tells us that wave speed depends both on the tension applied to the string and on its mass density.
The speed of a wave is consistent along the string as long as these factors remain unchanged.

Transverse waves, as seen in this example, travel perpendicular to the direction of the string's extension. This affects their speed and how they interact to form standing waves, which only occur at specific speeds that match the string's physical characteristics.
Understanding wave speed helps us predict how and when standing waves will form.

Linear Mass Density

Linear mass density \( \mu \) is a measure of the mass of the string per unit length. Its units can vary, but in this experiment it is given in grams per centimeter. A string's linear mass density impacts how it behaves under tension.

If we increase the linear mass density, for a fixed tension, the wave speed will decrease, since \( v = \sqrt{\frac{T}{\mu}} \).
This is because a more massive string will offer greater resistance to changes in motion.

• Heavier or thicker strings tend to have higher \( \mu \) values.
• Different strings can significantly affect the frequency and wavelength of waves.

In lab setups like this, testing different strings and observing the resultant wave patterns is crucial. It helps in understanding how mass distribution affects wave motion.

Tension in a String

Tension is a vital factor when examining wave motion on a string. The tension \( T \), given by the force applied to the string, is expressed as \( T = Mg \), where \( M \) is the mass hanging on the string, and \( g \) is the gravitational acceleration \( 9.8 \ m/s^2 \).

Tension influences wave speed, as seen in the equation \( v = \sqrt{\frac{T}{\mu}} \).
Higher tension results in a greater wave speed, which influences the patterns of standing waves. By varying \( M \), one can alter \( T \), allowing the observation of how different tensions alter wave behavior.

• High tension: Faster wave propagation.
• Low tension: Slower wave movement.

Understanding this concept is essential for creating precise standing waves, as these only form at particular tensions.

Transverse Waves

Transverse waves are specific types of waves where the displacement of the medium (the string, in this case) is perpendicular to the direction of wave propagation.
These waves are what oscillators produce in string experiments to form standing waves.

A standing wave can occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other.

• Nodes are points of zero amplitude.
• Antinodes are points of maximum amplitude.

The space between two consecutive nodes or antinodes is half of a wavelength in standing waves. Understanding transverse waves is crucial for interpreting wave patterns on a string, as these principles determine how and why waves behave the way they do under the experimental setup.