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A 121 -cm-long, \(4.0 \mathrm{g}\) string oscillates in its \(m=3\) mode with a frequency of \(180 \mathrm{Hz}\) and a maximum amplitude of \(5.0 \mathrm{mm}\) What are (a) the wavelength and (b) the tension in the string?

Short Answer

Expert verified
The wavelength is \(0.808\,m\) and the tension in the string is \(70.17\,N\)

Step by step solution

01

Calculate the Wavelength

Wavelength (\(λ\)) can be found since length of string (\(L\)) is related to wavelength as \(L=\frac{mλ}{2}\), where \(m\) is the mode number. In this case \(L=121\,cm = 1.21\,m\), \(m=3\). Solving for \(λ\), we find \(λ=\frac{2L}{m}\). Replacing the values gives us: \(λ=\frac{2*1.21}{3} = 0.808\,m\).
02

Calculate the Tension

The speed (\(v\)) of the oscillation can be calculated since \(v = λf\), where \(f\) is frequency, which is given as \(180\,Hz\). Therefore the speed is \(v = 0.808*180 = 145.4\,ms^{-1}\). The tension can be found by using the equation \(T = μv^2\), where \(μ\) is mass density. First calculate the mass density as \(μ = \frac{m}{L}\), whereby mass is \(4.0\,g = 0.004\,kg\), resulting in \(μ = \frac{0.004}{1.21} = 0.0033\,kg/m\). Then substitute \(μ\) and \(v\) into the equation for \(T\): \(T = 0.0033*(145.4)^2 = 70.17\,N\)

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Key Concepts

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

Wavelength Calculation
A wave on a string is an intriguing physical phenomenon that can be fully understood by determining its wavelength. The wavelength, denoted by \(λ\), is the distance between successive points that are phase-aligned along the string, such as peaks or troughs.
In our exercise, the string is oscillating in a specific mode, labelled as the third harmonic or \(m=3\). To find the wavelength when a string is fixed at both ends, we use the formula \(L=\frac{mλ}{2}\), where \(L\) is the length of the string. Breaking it down, it means the string accommodates multiple half-wavelengths within its length.
Plugging in the values, we find \[λ=\frac{2L}{m} = \frac{2 \times 1.21}{3} = 0.808\,m\]. This result tells us each complete wave fits 0.808 meters within the string's total length, providing insight into its oscillation pattern when vibrated in the third mode.
Tension in a String
Understanding the tension within a string helps explain how it reacts when set into motion. The tension, denoted as \(T\), determines how tightly the string is stretched and directly influences wave propagation speed.
We find tension by first determining the wave speed using the formula \(v = λf\), where \(f\) is the frequency. In this exercise, \(λ\) is 0.808 meters and \(f\) is 180 Hz, giving a speed \(v = 0.808 \times 180 = 145.4 \,ms^{-1}\).
To calculate tension, we need the linear mass density \(μ\), which is the mass per unit length of the string. Calculated as \(μ=\frac{m}{L}\) with mass \(m=0.004 \,kg\) and length \(L=1.21\,m\), it results in \(μ = 0.0033\,kg/m\). Finally, tension is given by \(T = μv^2\), leading to \(70.17 \,N\). Understanding this tension helps appreciate how physical forces in a string affect its movement.
Harmonic Frequency
When discussing waves on a string, the harmonic frequency is central. It signifies the number of vibrations or oscillations per second. In particular, each harmonic corresponds to a specific standing wave pattern.
For our example, the third harmonic frequency corresponds to the string's natural vibration mode, yielding complete waveform segments between endpoints. This frequency, given as 180 Hz, controls how quickly the wave oscillates, influencing characteristics like amplitude and wavelength.
Harmonic frequencies are inherently linked to their respective modes (modes defined by integer values \(m\)), with each mode increasing the complexity of the standing wave. Knowing the fundamental frequency—first harmonic—is crucial, as it establishes a base upon which higher harmonics build, hence helping to predict string behavior accurately.
String Oscillation
String oscillation describes how the string vibrates and forms standing waves. These oscillations occur when waves reflect back and forth along the string's surface. It helps visualize repetitive cycles, manifesting as nodes (points of no motion) and antinodes (points of maximum motion).
When a string is secured at both ends, like in the given problem, it forms distinct patterns, defined by harmonic modes. Each mode introduces a wave pattern of nodes and antinodes, illustrating regions of maximum swing and complete stillness.
Wave speed, wavelength, and frequency all synergistically influence oscillations. The harmonic mode highlights how the string divides into equal segments, exhibiting stable waveforms. Comprehending these oscillation principles aids in understanding how natural frequencies and external forces impact string behavior.

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Most popular questions from this chapter

Two sinusoidal waves with equal wavelengths travel along a string in opposite directions at \(3.0 \mathrm{m} / \mathrm{s}\). The time between two successive instants when the antinodes are at maximum height is \(0.25 \mathrm{s} .\) What is the wavelength?

A \(50-\mathrm{cm}\) -long wire with a mass of \(1.0 \mathrm{g}\) and a tension of \(440 \mathrm{N}\) passes across the open end of an open-closed tube of air. The wire, which is fixed at both ends, is bowed at the center so as to vibrate at its fundamental frequency and generate a sound wave. Then the tube length is adjusted until the fundamental frequency of the tube is heard. What is the length of the tube? Assume \(v_{\text {sound }}=340 \mathrm{m} / \mathrm{s}\).

Standing waves on a \(1.0-\) m-long string that is fixed at both ends are seen at successive frequencies of \(24 \mathrm{Hz}\) and \(36 \mathrm{Hz}\) a. What are the fundamental frequency and the wave speed? b. Draw the standing-wave pattern when the string oscillates at \(36 \mathrm{Hz}\)

Il Example 21.10 showed that a 92 -nm-thick coating of \(\mathrm{MgF}_{2}\) \((n=1.39)\) on glass acts as an antireflection coating for light with a wavelength of \(510 \mathrm{nm}\). Without the coating, the intensity of reflected light is \(I_{0}=c a^{2},\) where \(a\) is the amplitude of the reflected light wave and \(c\) is an unknown proportionality constant. a. Let \(I_{\lambda}\) be the intensity of light reflected from the coated glass at wavelength \(\lambda\). Find an expression for the ratio \(I_{\lambda} / I_{0}\) as a function of the wavelength \(\lambda\). This ratio is the reflection intensity from the coated glass relative to the reflection intensity from uncoated glass. A ratio less than 1 indicates that the coating is reducing the reflection intensity. Hint: The amplitude of the superposition of two waves depends on the phase difference between the waves. Although not entirely accurate, assume that both reflected waves have amplitude \(a\). b. Evaluate \(I_{A} / I_{0}\) at \(\lambda=400,450,500,550,600,650,\) and \(700 \mathrm{nm} .\) This spans the range of visible light. c. Draw a graph of \(I_{\lambda} / I_{0}\) versus \(\lambda\).

What is the thinnest film of \(\mathrm{MgF}_{2}(n=1.39)\) on glass that produces a strong reflection for orange light with a wavelength of \(600 \mathrm{nm} ?\)

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