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Chapter 15: Problem 19

The displacement from equilibrium caused by a wave on a string is given by \(y(x, t)=(-0.00200 \mathrm{~m}) \sin \left[\left(40.0 \mathrm{~m}^{-1}\right) x-\right.\) \(\left.\left(800 . \mathrm{s}^{-1}\right) t\right] .\) For this wave, what are the (a) amplitude, (b) number of waves in \(1.00 \mathrm{~m},\) (c) number of complete cycles in \(1.00 \mathrm{~s},\) (d) wavelength, and (e) speed?

Short Answer

Expert verified
Answer: The amplitude is \(0.00200m\), the number of waves in 1.00m is 40, there are approximately 127.32 complete cycles in 1.00s, the wavelength is approximately \(0.15708m\), and the wave speed is approximately \(20m/s\).

Step by step solution

01

(a) Amplitude

To find the amplitude, compare the given wave equation with the general equation of a sinusoidal wave. The amplitude \(A\) is the coefficient of the sine function. In this case, it is \(-0.00200 m\). The amplitude is always a positive value, so the amplitude of this wave is \(0.00200m\).
02

(b) Number of Waves in 1.00m

To find the number of waves in 1.00m, we need the wave number \(k\). The wave number is found by looking at the coefficient of the \(x\) term in the sine function. In this case, the wave number \(k = 40.0 m^{-1}\). The number of waves in 1.00m is equal to the wave number, which in this case is 40.
03

(c) Number of Complete Cycles in 1.00s

To find the number of complete cycles in 1.00s, we need the angular frequency \(\omega\). The angular frequency is found by looking at the coefficient of the \(t\) term in the sine function. In this case, the angular frequency \(\omega = 800 s^{-1}\). To find the number of cycles, we need to convert angular frequency to frequency (\(f\)). The relationship between angular frequency and frequency is given by \(\omega = 2\pi f\). Therefore, we have \(f = \frac{\omega}{2\pi} = \frac{800}{2\pi} \approx 127.32 Hz\). This means there are 127.32 complete cycles in 1.00s.
04

(d) Wavelength

To find the wavelength, we need to use the relationship between wavelength and wave number, which is given by \(\lambda = \frac{2\pi}{k}\). Using the given wave number \(k = 40.0 m^{-1}\), we can find the wavelength as \(\lambda = \frac{2\pi}{40} = \frac{\pi}{20} \approx 0.15708m\).
05

(e) Wave Speed

To find the wave speed, we use the relationship between speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) given by \(v = f\lambda\). We have the frequency \(f \approx 127.32 Hz\) and wavelength \(\lambda \approx 0.15708m\). So, the wave speed is \(v \approx 127.32 \times 0.15708 \approx 20m/s\).

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Most popular questions from this chapter

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 \(t=0\).

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?

A guitar string with a mass of \(10.0 \mathrm{~g}\) is \(1.00 \mathrm{~m}\) long and attached to the guitar at two points separated by \(65.0 \mathrm{~cm} .\) a) What is the frequency of the first harmonic of this string when it is placed under a tension of \(81.0 \mathrm{~N} ?\) b) If the guitar string is replaced by a heavier one that has a mass of \(16.0 \mathrm{~g}\) and is \(1.00 \mathrm{~m}\) long, what is the frequency of the replacement string's first harmonic?

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