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Chapter 11: Problem 48

A harpsichord string of length \(1.50 \mathrm{m}\) and linear mass density $25.0 \mathrm{mg} / \mathrm{m}\( vibrates at a (fundamental) frequency of \)450.0 \mathrm{Hz}$. (a) What is the speed of the transverse string waves? (b) What is the tension? (c) What are the wavelength and frequency of the sound wave in air produced by vibration of the string? (The speed of sound in air at room temperature is \(340 \mathrm{m} / \mathrm{s} .\) )

Short Answer

Expert verified
Answer: The speed of the transverse string waves is 1350 m/s, the tension in the string is 45,562.50 N, and the wavelength and frequency of the sound wave in air are approximately 0.7556 m and 450 Hz, respectively.

Step by step solution

01

Find the speed of the transverse string waves

To find the speed of the transverse string waves (v), we will use the wave speed formula: v = fλ where: f is the (fundamental) frequency, given as 450 Hz λ is the wavelength of the string Given that the string is 1.50 m long, when it vibrates at its fundamental frequency, only half a wave fits into the length, which means the wavelength is twice the length of the string: λ = 2L = 2 * 1.50 m = 3.00 m Now we can calculate the wave speed (v): v = fλ = (450 Hz)(3.00 m) = 1350 m/s.
02

Find the tension in the string

To find the tension (T) in the string, we can use the formula for the wave speed of a vibrating string: v = sqrt(T/μ) where: μ is the linear mass density of the string, given as 0.025 kg/m (converted from mg/m) We can rearrange the formula to solve for T: T = μv^2 = (0.025 kg/m)(1350 m/s)^2 = 45,562.50 N
03

Find the wavelength and frequency of the sound wave in air

Now, we need to find the wavelength (λ_air) and frequency (f_air) of the sound wave in air produced by the vibration of the string. We are given that the speed of sound in air is 340 m/s. First, we find the wavelength by using the formula for the speed of sound in air (v_air = f_airλ_air): λ_air = v_air/f λ_air = (340 m/s)/(450 Hz) ≈ 0.7556 m Since the frequency of the sound wave in air is the same as the frequency of the string wave, we have: f_air = f = 450 Hz So, the wavelength and frequency of the sound wave in air are approximately 0.7556 m and 450 Hz, respectively.

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Most popular questions from this chapter

Two speakers spaced a distance \(1.5 \mathrm{m}\) apart emit coherent sound waves at a frequency of \(680 \mathrm{Hz}\) in all directions. The waves start out in phase with each other. A listener walks in a circle of radius greater than one meter centered on the midpoint of the two speakers. At how many points does the listener observe destructive interference? The listener and the speakers are all in the same horizontal plane and the speed of sound is \(340 \mathrm{m} / \mathrm{s} .\) [Hint: Start with a diagram; then determine the maximum path difference between the two waves at points on the circle. I Experiments like this must be done in a special room so that reflections are negligible.
What is the speed of the wave represented by \(y(x, t)=\) $A \sin (k x-\omega t),\( where \)k=6.0 \mathrm{rad} / \mathrm{cm}\( and \)\omega=5.0 \mathrm{rad} / \mathrm{s} ?$
(a) Plot a graph for $$ y(x, t)=(4.0 \mathrm{cm}) \sin [(378 \mathrm{rad} / \mathrm{s}) t-(314 \mathrm{rad} / \mathrm{cm}) x] $$ versus \(x\) at \(t=0\) and at \(t=\frac{1}{480} \mathrm{s}\). From the plots determine the amplitude, wavelength, and speed of the wave. (b) For the same function, plot a graph of \(y(x, t)\) versus \(t\) at \(x=0\) and find the period of the vibration. Show that \(\lambda=v T.\)
Show that the amplitudes of the graphs you made in Problem 74 satisfy the equation \(A^{\prime}=2 A \cos (\omega t),\) where \(A^{\prime}\) is the amplitude of the wave you plotted and \(A\) is \(5.0 \mathrm{cm},\) the amplitude of the waves that were added together.
When the string of a guitar is pressed against a fret, the shortened string vibrates at a frequency \(5.95 \%\) higher than when the previous fret is pressed. If the length of the part of the string that is free to vibrate is \(64.8 \mathrm{cm},\) how far from one end of the string are the first three frets located?
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