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Chapter 15: Problem 61

Consider a guitar string stretching \(80.0 \mathrm{~cm}\) between its anchored ends. The string is tuned to play middle \(\mathrm{C},\) with a frequency of \(256 \mathrm{~Hz}\), when oscillating in its fundamental mode, that is, with one antinode between the ends. If the string is displaced \(2.00 \mathrm{~mm}\) at its midpoint and released to produce this note, what are the wave speed, \(v\), and the maximum speed, \(V_{\text {max }}\), of the midpoint of the string?

Short Answer

Expert verified
Answer: The wave speed is approximately 409.6 m/s, and the maximum speed of the midpoint of the string is approximately 10.05 m/s.

Step by step solution

01

Find the wavelength of the fundamental mode

In the fundamental mode, the length of the string is equal to half the wavelength. So, to find the wavelength \(λ\), we can use the relationship: \(λ = 2L\) Where \(L\) is the length of the string. We are given the length of the string, \(L = 80.0 \mathrm{~cm}\), so we can calculate the wavelength: \(λ = 2 \times 80.0 \mathrm{~cm} = 160.0 \mathrm{~cm} = 1.6 \mathrm{~m}\)
02

Calculate the wave speed

Now that we have the wavelength, we can calculate the wave speed using the formula: \(v = fλ\) We are given the frequency \(f = 256 \mathrm{~Hz}\) and we have found the wavelength \(λ = 1.6 \mathrm{~m}\), so we can calculate the wave speed: \(v = 256 \mathrm{~Hz} \times 1.6 \mathrm{~m} = 409.6 \mathrm{~m/s}\)
03

Calculate the angular frequency

To find the maximum speed of the midpoint, we need the angular frequency \(ω\). We can find it using the formula: \(ω = 2πf\) Where \(f\) is the frequency. We are given \(f = 256 \mathrm{~Hz}\), so we can calculate the angular frequency: \(ω = 2π \times 256 \mathrm{~Hz} = 512π \mathrm{~rad/s}\)
04

Find the maximum speed of the midpoint

Now we can find the maximum speed of the midpoint using the formula: \(V_{max} = Aω\) Where \(A\) is the amplitude and \(ω\) is the angular frequency. We are given the displacement of the midpoint, \(A = 2.00 \mathrm{~mm} = 0.002 \mathrm{~m}\), and we have found the angular frequency, \(ω = 512π \mathrm{~rad/s}\). We can calculate the maximum speed: \(V_{max} = 0.002 \mathrm{~m} \times 512π \mathrm{~rad/s} \approx 3.200π \mathrm{~m/s} \approx 10.05 \mathrm{~m/s}\) To summarize the results: - The wave speed, \(v\), is approximately \(409.6 \mathrm{~m/s}\). - The maximum speed, \(V_{max}\), of the midpoint of the string is approximately \(10.05 \mathrm{~m/s}\).

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Most popular questions from this chapter

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?

a) Starting from the general wave equation (equation 15.9 ), prove through direct derivation that the Gaussian wave packet described by the equation \(y(x, t)=(5.00 m) e^{-0.1(x-5 t)^{2}}\) is indeed a traveling wave (that it satisfies the differential wave equation). b) If \(x\) is specified in meters and \(t\) in seconds, determine the speed of this wave. On a single graph, plot this wave as a function of \(x\) at \(t=0, t=1.00 \mathrm{~s}, t=2.00 \mathrm{~s},\) and \(t=3.00 \mathrm{~s}\) c) More generally, prove that any function \(f(x, t)\) that depends on \(x\) and \(t\) through a combined variable \(x \pm v t\) is a solution of the wave equation, irrespective of the specific form of the function \(f\)

Which of the following transverse waves has the greatest power? a) a wave with velocity \(v\), amplitude \(A\), and frequency \(f\) b) a wave of velocity \(v\), amplitude \(2 A\), and frequency \(f / 2\) c) a wave of velocity \(2 v\), amplitude \(A / 2\), and frequency \(f\) d) a wave of velocity \(2 v\), amplitude \(A\), and frequency \(f / 2\) e) a wave of velocity \(v\), amplitude \(A / 2\), and frequency \(2 f\)

Why do circular water waves on the surface of a pond decrease in amplitude as they travel away from the source?

Hiking in the mountains, you shout "hey," wait \(2.00 \mathrm{~s}\) and shout again. What is the distance between the sound waves you cause? If you hear the first echo after \(5.00 \mathrm{~s}\), what is the distance between you and the point where your voice hit a mountain?
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