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Three Lowest Frequencies What are the three lowest frequencies for standing waves on a wire \(10.0 \mathrm{~m}\) long having a mass of \(100 \mathrm{~g}\), which is stretched under a tension of \(250 \mathrm{~N} ?\)

Short Answer

Expert verified
The three lowest frequencies are 7.91 Hz, 15.81 Hz, and 23.72 Hz.

Step by step solution

01

- Calculate the Linear Mass Density

First, find the linear mass density, \(\rho\), of the wire. This is done by dividing the mass of the wire by its length. The given mass is 100 g or 0.1 kg and the length is 10.0 m. So, \( \rho = \frac{0.1 \text{ kg}}{10.0 \text{ m}} = 0.01 \text{ kg/m} \).
02

- Find the Wave Speed

Next, calculate the wave speed, \(v\), on the wire using the formula \( v = \sqrt{\frac{T}{\rho}} \), where \( T \) is the tension. Substituting the given tension \(T = 250 \text{ N}\) and linear mass density \(\rho = 0.01 \text{ kg/m}\), we get \( v = \sqrt{\frac{250}{0.01}} = \sqrt{25000} = 158.11 \text{ m/s} \).
03

- Determine the Fundamental Frequency

The fundamental frequency, \( f_1 \, \), is given by \( \frac{v}{2L} \.\) Given \( L = 10 \text{ m}\), substituting \( v\), we get \( f_1 = \frac{158.11}{2 \times \ 10} = 7.91 \text{ Hz} \).
04

- Determine the Second Harmonic

The frequency of the second harmonic \( f_2 \, \), can be found using \( f_2 = \frac{2v}{2L} = \frac{v}{L} \). So, \( f_2 = \frac{158.11}{10} = 15.81 \text{ Hz} \).
05

- Determine the Third Harmonic

Similarly, the frequency of the third harmonic \( f_3 \, \), is given by \( f_3 = \frac{3v}{2L} \), which simplifies to \( \frac{3 \ \times \ 158.11}{2 \times \ 10} = 23.72 \text{ Hz} \).

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Key Concepts

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

linear mass density
The linear mass density, often denoted as \( \rho \), is a measure of how much mass is distributed along the length of a wire or string. It is calculated using the formula:
\[ \rho = \frac{m}{L} \]
where \( m \) is the mass and \( L \) is the length. In the exercise, the wire has a mass of 100 grams, which is equivalent to 0.1 kilograms, and a length of 10 meters:
\[ \rho = \frac{0.1 \text{ kg}}{10 \text{ m}} = 0.01 \text{ kg/m} \]
This means each meter of the wire has a mass of 0.01 kilograms. A lower linear mass density usually means the wire is lighter, which impacts the speed at which waves can travel through it.
wave speed calculation
The speed at which waves travel through a medium, such as a wire, depends on two factors: the tension force \( T \) applied to the wire and its linear mass density \( \rho \). The formula to calculate wave speed \( v \) is given by:
\[ v = \sqrt{\frac{T}{\rho}} \]
Substituting the given tension \( T = 250 \text{ N} \) and \( \rho = 0.01 \text{ kg/m} \), we find: \[ v = \sqrt{\frac{250}{0.01}} = \sqrt{25000} = 158.11 \text{ m/s} \]
The wave speed is critical because it directly affects the frequencies of standing waves that can form in the wire, with higher speeds resulting in higher frequencies.
harmonic frequencies
Harmonic frequencies are specific standing wave frequencies that occur at natural resonance points within the wire. The fundamental frequency \( f_1 \) is the lowest frequency at which a standing wave can form and is calculated using:
\[ f_1 = \frac{v}{2L} \]
For a wave speed \( v = 158.11 \text{ m/s} \) and wire length \( L = 10 \text{ m} \), we get:
\[ f_1 = \frac{158.11}{2 \times 10} = 7.91 \text{ Hz} \]
The second harmonic \( f_2 \) is twice the fundamental frequency:
\[ f_2 = \frac{2v}{2L} = \frac{v}{L} = 15.81 \text{ Hz} \]
The third harmonic \( f_3 \) is three times the fundamental frequency:
\[ f_3 = \frac{3v}{2L} = 23.72 \text{ Hz} \]
Each harmonic frequency corresponds to a different pattern of standing waves on the wire, with increasing frequencies having more nodes and antinodes along the length of the wire.

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Most popular questions from this chapter

Tuning Fork Oscillation of a \(600 \mathrm{~Hz}\) tuning fork sets up standing waves in a string clamped at both ends. The wave speed for the string is \(400 \mathrm{~m} / \mathrm{s}\). The standing wave has four loops and an amplitude of \(2.0 \mathrm{~mm}\). (a) What is the length of the string? (b) Write an equation for the displacement of the string as a function of position and time.

Displacement of Particles A sinusoidal wave is traveling on a string with speed \(40 \mathrm{~cm} / \mathrm{s}\). The displacement of the particles of the string at \(x=10 \mathrm{~cm}\) is found to vary with time according to the equation \(y(x, t)=(5.0 \mathrm{~cm}) \sin [1.0 \mathrm{rad} / \mathrm{cm}-(4.0 \mathrm{rad} / \mathrm{s}) t] .\) The linear density of the string is \(4.0 \mathrm{~g} / \mathrm{cm} .\) What are the (a) frequency and (b) wavelength of the wave? (c) Write the general equation giving the transverse displacement of the particles of the string as a function of position and time. (d) Calculate the tension in the string.

Show That Show that $$ \begin{array}{ll} y(x, t)=Y \sin k(x-v t), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-f t\right) \\ y(x, t)=Y \sin \omega\left(\frac{x}{v}-t\right), & y(x, t)=Y \sin 2 \pi\left(\frac{x}{\lambda}-\frac{t}{T}\right) \end{array} $$ are all equivalent to \(y(x, t)=Y \sin (k x-\omega t)\).

Linear Density The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y(x, t)=(0.021 \mathrm{~m}) \sin [(2.0 \mathrm{rad} / \mathrm{m}) x+(30 \mathrm{rad} / \mathrm{s}) t] $$ What is (a) the wave speed and (b) the tension in the string?

Resonant Frequencies A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

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