/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A physics student suspends a 1 k... [FREE SOLUTION] | 91影视

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A physics student suspends a 1 kg mass from a pulley using a wire as shown in Figure \(12.39 .\) The pulley allows her to easily change the length of the wire. In order to determine the speed of a transverse wave on the wire, she measures the frequency of the fundamental mode as a function of the length of the wire \(L\). The table shows her results: $$\begin{array}{lc}\hline L(\mathrm{~m}) & f(\mathrm{H} z) \\\\\hline 1.0 & 48 \\\0.5 & 98 \\\0.3 & 160 \\\0.2 & 249 \\\\\hline\end{array}$$ Make two plots of these data. First plot the frequency as a function of \(L_{\text {, and then plot it as a function of } 1 / L \text { . Which graph shows a }}\) linear relationship? Draw a best-fit line through the linear graph's data, and use the slope of this line to determine the wave speed.

Short Answer

Expert verified
The linear plot is of frequency vs \( 1/L \), and the wave speed is 100.5 m/s.

Step by step solution

01

Understand the Relationship

The frequency of the fundamental mode of a wave on a string is related to the length of the string. The frequency \( f \) is inversely related to the length \( L \) of the string, meaning \( f \propto \frac{1}{L} \). This implies that plotting \( f \) against \( \frac{1}{L} \) should yield a linear graph.
02

Prepare the Data

Calculate the reciprocal of lengths \( \frac{1}{L} \) for each given value of \( L \): - For \( L = 1.0 \) m, \( \frac{1}{L} = 1.0 \) m鈦宦- For \( L = 0.5 \) m, \( \frac{1}{L} = 2.0 \) m鈦宦- For \( L = 0.3 \) m, \( \frac{1}{L} 鈮 3.33 \) m鈦宦- For \( L = 0.2 \) m, \( \frac{1}{L} = 5.0 \) m鈦宦
03

Plot Frequency vs Length

Plot the given frequency \( f \) values against the wire length \( L \): - (1.0, 48)- (0.5, 98)- (0.3, 160)- (0.2, 249)This plot typically does not show a linear relationship given the data set.
04

Plot Frequency vs Reciprocal Length

Plot the frequency \( f \) against \( \frac{1}{L} \) values:- (1.0, 48)- (2.0, 98)- (3.33, 160)- (5.0, 249)This plot should produce approximately a straight line, indicating a linear relationship.
05

Determine the Slope of the Linear Plot

Fit a best-fit line through the points on the frequency versus \( \frac{1}{L} \) plot.Calculate the slope \( m \) of this line. Use points \((x_1, y_1) = (1.0, 48)\) and \((x_2, y_2) = (5.0, 249)\) to estimate:\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{249 - 48}{5 - 1} = 50.25\] Hz m.
06

Calculate the Wave Speed

The wave speed \( v \) is determined using the relationship \( v = 2L \times f \). Rearranging terms for a string vibrating in its fundamental mode gives \( v = 2 \times \text{slope}\). The speed of the wave on the wire is:\( v = 2 \times 50.25 = 100.5 \) m/s.

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Key Concepts

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

Fundamental Frequency
The fundamental frequency, often referred to as the first harmonic, is the lowest frequency at which a system naturally vibrates. It is a crucial concept when analyzing waves on strings, especially in physics experiments like the one described in the exercise. When a string is plucked or struck, it begins vibrating at this fundamental frequency, producing a pitch or tone. The fundamental frequency depends on several factors: the length of the string, its tension, and its mass per unit length. In simpler terms, altering any of these factors will change the fundamental frequency. For example, increasing the tension will raise the frequency, while lengthening the string will lower it.
String Length and Wave Relationship
Understanding the interplay between string length and wave behavior is essential in wave mechanics. The length of a string is inversely related to the frequency of the wave it produces. This means that as the length of the string increases, the frequency of the wave decreases, and vice versa. A practical view of this can be observed in musical instruments. A guitarist shortens the string by pressing down on a fret to increase the pitch, effectively raising the wave frequency. The mathematical expression of this relationship in the exercise is shown as \( f \propto \frac{1}{L} \), where \( f \) is the frequency, and \( L \) is the string length.
Inverse Proportionality
Inverse proportionality is a mathematical relationship where one quantity increases while another decreases. In the given exercise, the frequency and the length of the string are inversely proportional. This relationship can be expressed with \( f \propto \frac{1}{L} \), which implies that as the length of the string declines, the frequency escalates. This concept is crucial in physics and engineering because it allows us to predict how changes in one variable will affect another. If students create a graph plotting \( f \) against \( \frac{1}{L} \), the data should reveal a straight line, highlighting the linear nature facilitated by inverse proportionality.
Best-Fit Line
Drawing a best-fit line is a fundamental technique in data analysis, used to identify the underlying trend of a data set. In the context of the exercise, plotting the frequency against the reciprocal of length and fitting a best-fit line helps to visualize the linear relationship driven by inverse proportionality. The slope of this line is particularly important as it is used to estimate the wave speed. By selecting appropriate data points, students can calculate the slope using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). This slope is then doubled to find the wave speed, linking the graphical analysis to a physical characteristic of the wave, enhancing comprehension of both mathematical and physical applications.

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Most popular questions from this chapter

The sound source of a ship's sonar system operates at a frequency of \(22.0 \mathrm{kHz}\). The speed of sound in water (assumed to be at a uniform \(20^{\circ} \mathrm{C}\) ) is \(1482 \mathrm{~m} / \mathrm{s}\). (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at \(4.95 \mathrm{~m} / \mathrm{s} ?\) The ship is at rest in the water.

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is \(344 \mathrm{~m} / \mathrm{s},\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is \(14.0 \mathrm{~cm},\) what is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

Two train whistles, \(A\) and \(B\), each have a frequency of \(392 \mathrm{~Hz}\). \(A\) is stationary and \(B\) is moving toward the right (away from \(A\) ) at a speed of \(35.0 \mathrm{~m} / \mathrm{s}\). A listener is between the two whistles and is moving toward the right with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). (See Figure \(12.43 .)\) (a) What is the frequency from \(A\) as heard by the listener? (b) What is the frequency from \(B\) as heard by the listener? (c) What is the beat frequency detected by the listener?

The range of audible frequencies is from about \(20 \mathrm{~Hz}\) to \(20.000 \mathrm{~Hz}\). What is the range of the wavelengths of audible sound in air? (b) Visible light. The range of visible light extends from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\). What is the range of visible frequencies of light? (c) Brain surgery. Surgeons can remove brain tumors by using a cavitron ultrasonic surgical aspirator, which produces sound waves of frequency \(23 \mathrm{kHz}\). What is the wavelength of these waves in air? (d) Sound in the body. What would be the wavelength of the sound in part (c) in bodily fluids in which the speed of sound is \(1480 \mathrm{~m} / \mathrm{s},\) but the frequency is unchanged?

You are trying to overhear a juicy conversation, but from your distance of \(15.0 \mathrm{~m}\), it sounds like only an average whisper of \(20.0 \mathrm{~dB}\). So you decide to move closer to give the conversation a sound level of \(60.0 \mathrm{~dB}\) instead. How close should you come?

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