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Chapter 10: Problem 1511

Standing waves are produced by the superposition of two waves $y_{1}=0.05 \sin (3 \pi t-2 x)\( and \)y_{2}=0.05 \sin (3 \pi t+2 x)$ where \(\mathrm{x}\) and \(\mathrm{y}\) are in meters and \(\mathrm{t}\) is in seconds. The speed (in \(\mathrm{ms}-1\) ) of each wave is \(\ldots \ldots\) (A) \(1.5\) (B) \(3.0\) (C) \(3 \pi / 2\) (D) \(3 \pi\)

Short Answer

Expert verified
The speed of each wave is \(v = \frac{3 \pi}{2}\) m/s.

Step by step solution

01

Identify wavelength and frequency from the equations

We are given the two waves: \(y_{1}=0.05 \sin (3 \pi t-2 x)\) and \(y_{2}=0.05 \sin (3 \pi t+2 x)\). These are sinusoidal waves described by the form \( y = A \sin (\omega t \pm kx) \), where A is the amplitude, ω is the angular frequency, k is the wave number, and ± depends on the direction of each wave. For both waves, we are given \(\omega = 3 \pi\), k = 2. Next, we will find the lambda (wavelength) and frequency (f).
02

Calculate wavelength

The wavelength (lambda) is related to the wave number as follows: \( k = \frac{2 \pi}{\lambda} \) We know k = 2, therefore the wavelength is: \( \lambda = \frac{2 \pi}{k} = \frac{2 \pi}{2} = \pi \) meters.
03

Calculate frequency

Frequency is related to the angular frequency as follows: \( f = \frac{\omega}{2 \pi} \) We know \(\omega = 3 \pi\), and therefore the frequency is: \( f = \frac{3 \pi}{2 \pi} = \frac{3}{2} \) Hz.
04

Determine wave speed

Now we have the wavelength (\(\lambda = \pi\)) and the frequency (\(f = \frac{3}{2}\)). We will use the wave speed equation (\( v = \lambda f\)) to calculate the speed of the wave: \( v = (\pi) \left(\frac{3}{2}\right) = \frac{3 \pi}{2} \) meters per second. The answer is choice (C): \(v = \frac{3 \pi}{2}\) m/s.

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Most popular questions from this chapter

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