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Chapter 15: Problem 39

A 2.00 -m-long string of mass \(10.0 \mathrm{~g}\) is clamped at both ends. The tension in the string is \(150 \mathrm{~N}\). a) What is the speed of a wave on this string? b) The string is plucked so that it oscillates. What is the wavelength and frequency of the resulting wave if it produces a standing wave with two antinodes?

Short Answer

Expert verified
Based on the given information, the speed of the wave on the string is 173 m/s, and the wavelength and frequency of the standing wave with two antinodes are 2 meters and 86.5 Hz, respectively.

Step by step solution

01

Calculate the wave speed

To find the speed of a wave on the string, we can use the formula: Wave speed (v) = \(\sqrt{\frac{T}{\mu}}\) where T is the tension in the string and \(\mu\) is the linear mass density of the string, defined as the mass per unit length. First, we need to find the linear mass density. This can be calculated using: \(\mu\) = \(\frac{m}{L}\) where m is the mass of the string, and L is the length of the string.
02

Calculate the linear mass density

We are given the mass of the string is 10g, and it is 2 meters long. So the linear mass density can be calculated as: \(\mu\) = \(\frac{10 \text{g}}{2 \text{m}}\) = \(\frac{10 \times 10^{-3} \text{kg}}{2 \text{m}}\) = \(5\times10^{-3} \frac{\text{kg}}{\text{m}}\)
03

Calculate the wave speed

Now that we have the linear mass density, we can plug the values of tension and \(\mu\) into the wave speed formula: v = \(\sqrt{\frac{150 \, \text{N}}{5\times10^{-3} \, \frac{\text{kg}}{\text{m}}}}\) v = \(\sqrt{30,000 \, \frac{\text{m}^2}{\text{s}^2}} = 173 \, \frac{\text{m}}{\text{s}}\) The wave speed on this string is 173 m/s.
04

Determine the wavelength of the standing wave

Now that we have the wave speed, we can move on to finding the wavelength of the standing wave with two antinodes. For a standing wave with two antinodes, the string length is equal to one full wavelength. So, the wavelength (\(\lambda\)) is: \(\lambda = 2 \text{m}\)
05

Determine the frequency of the standing wave

Finally, we can calculate the frequency of the standing wave using the wave speed, v, and the wavelength, \(\lambda\): Frequency (f) = \(\frac{v}{\lambda}\) f = \(\frac{173 \, \frac{\text{m}}{\text{s}}}{2 \, \text{m}}\) = \(86.5 \text{ Hz}\) The frequency of the standing wave with two antinodes is 86.5 Hz. In summary, the speed of a wave on the string is 173 m/s, and the wavelength and frequency of the resulting standing wave with two antinodes are 2 meters and 86.5 Hz, respectively.

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Most popular questions from this chapter

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(T=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different frequencies \((\omega=3000\). rad/s and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s}\) ) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

A string is \(35.0 \mathrm{~cm}\) long and has a mass per unit length of \(5.51 \cdot 10^{-4} \mathrm{~kg} / \mathrm{m}\). What tension must be applied to the string so that it vibrates at the fundamental frequency of \(660 \mathrm{Hz?}\)

Two steel wires are stretched under the same tension. The first wire has a diameter of \(0.500 \mathrm{~mm}\), and the second wire has a diameter of \(1.00 \mathrm{~mm}\). If the speed of waves traveling along the first wire is \(50.0 \mathrm{~m} / \mathrm{s}\), what is the speed of waves traveling along the second wire?

A wave travels along a string in the positive \(x\) -direction at \(30.0 \mathrm{~m} / \mathrm{s}\). The frequency of the wave is \(50.0 \mathrm{~Hz}\). At \(x=0\) and \(t=0,\) the wave velocity is \(2.50 \mathrm{~m} / \mathrm{s}\) and the vertical displacement is \(y=4.00 \mathrm{~mm} .\) Write the function \(y(x, t)\) for the wave

A ping-pong ball is floating in the middle of a lake and waves begin to propagate on the surface. Can you think of a situation in which the ball remains stationary? Can you think of a situation involving a single wave on the lake in which the ball remains stationary?
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