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Chapter 15: Problem 53

What is the wave speed along a brass wire with a radius of \(0.500 \mathrm{~mm}\) stretched at a tension of \(125 \mathrm{~N}\) ? The density of brass is \(8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
Answer: The wave speed along the brass wire is approximately 54.17 m/s.

Step by step solution

01

Calculate the cross-sectional area

Given the radius of the wire \(r = 0.500 \mathrm{~mm}\), we can find the cross-sectional area \(A\) using the formula for the area of a circle: \(A = \pi r^2\). So, $$A = \pi (0.500 \times 10^{-3} \mathrm{~m})^2 \approx 7.854 \times 10^{-7} \mathrm{~m^2}$$
02

Calculate the volume of one meter of the wire

To find the volume of one meter of the wire, we need to multiply the cross-sectional area by the length of the wire. Since we are considering one meter of the wire, we simply use the area we calculated in Step 1: $$V = A \times 1\mathrm{~m} = 7.854 \times 10^{-7} \mathrm{~m^3}$$
03

Calculate the mass of one meter of the wire

Now that we have the volume of one meter of the wire, we can find its mass using the density of brass, \(\rho = 8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The formula for mass is \(m = \rho V\): $$m = (8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3})(7.854 \times 10^{-7} \mathrm{~m^3}) \approx 6.753 \times 10^{-3} \mathrm{~kg}$$
04

Calculate the linear mass density of the wire

The linear mass density \(\mu\) is the mass per unit length, in this case, the mass of one meter of the wire. We already calculated the mass of one meter of the wire in Step 3, so we can directly use that value: $$\mu = \frac{m}{1 \mathrm{~m}} = 6.753 \times 10^{-3} \mathrm{~kg/m}$$
05

Find the wave speed along the wire

With the linear mass density \(\mu\) and the given tension \(T = 125 \mathrm{~N}\), we can now use the formula for wave speed on a string to find the wave speed \(v\): $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{125 \mathrm{~N}}{6.753 \times 10^{-3} \mathrm{~kg/m}}} \approx 54.17 \mathrm{~m/s}$$ Thus, the wave speed along the brass wire is approximately \(54.17 \mathrm{~m/s}\).

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Most popular questions from this chapter

A sinusoidal wave traveling in the positive \(x\) -direction has a wavelength of \(12 \mathrm{~cm},\) a frequency of \(10.0 \mathrm{~Hz},\) and an amplitude of \(10.0 \mathrm{~cm}\). The part of the wave that is at the origin at \(t=0\) has a vertical displacement of \(5.00 \mathrm{~cm} .\) For this wave, determine the a) wave number, d) speed, b) period, e) phase angle, and c) angular frequency, f) equation of motion.

Write the equation for a sinusoidal wave propagating in the negative \(x\) -direction with a speed of \(120 . \mathrm{m} / \mathrm{s}\), if a particle in the medium in which the wave is moving is observed to swing back and forth through a \(6.00-\mathrm{cm}\) range in \(4.00 \mathrm{~s}\). Assume that \(t=0\) is taken to be the instant when the particle is at \(y=0\) and that the particle moves in the positive \(y\) -direction immediately after \(t=0\).

If two traveling waves have the same wavelength, frequency, and amplitude and are added appropriately, the result is a standing wave. Is it possible to combine two standing waves in some way to give a traveling wave?

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.75 x-40 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10 \mathrm{~g} / \mathrm{m}\), determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\) (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.

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 \(660 \mathrm{Hz?}\)
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