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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

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.

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

A wave on a string has a wave function given by $$ y(x, t)=(0.0200 \mathrm{~m}) \sin \left[\left(6.35 \mathrm{~m}^{-1}\right) x+\left(2.63 \mathrm{~s}^{-1}\right) t\right] $$ a) What is the amplitude of the wave? b) What is the period of the wave? c) What is the wavelength of the wave? d) What is the speed of the wave? e) In which direction does the wave travel?

Fans at a local football stadium are so excited that their team is winning that they start "the wave" in celebration. Which of the following four statements is (are) true? I. This wave is a traveling wave. II. This wave is a transverse wave. III. This wave is a longitudinal wave. IV. This wave is a combination of a longitudinal wave and a transverse wave. a) I and II c) III only e) I and III b) II only d) I and IV

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?
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