/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The amplitude of a wave disturba... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 23

The amplitude of a wave disturbance propagating in the positive \(y\)-direction is given by \(y=\frac{1}{1+x^{2}}\) at \(t=\) 0 and \(y=\frac{1}{\left[1+(x-1)^{2}\right]}\) at \(t=2\) second, where \(x\) and \(y\) are in \(m\). If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave? (A) \(1 \mathrm{~m} / \mathrm{s}\) (B) \(1.5 \mathrm{~m} / \mathrm{s}\) (C) \(0.5 \mathrm{~m} / \mathrm{s}\) (D) \(2 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity of the wave is 0.5 m/s. The correct answer is (C) \(0.5 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Identify the position of the peak in each expression

The peak of each expression representing the wave amplitude is located where the value of y is maximum. Since the denominators of both expressions are positive, the maximum value of y occurs when the denominator is at its smallest value. This happens when x=0 for the first expression, and x=1 for the second expression.
02

Calculate the distance travelled by the peak

Since the peak of the wave is found at x=0 at t=0 and at x=1 at t=2 seconds, the peak has travelled from x=0 to x=1 in 2 seconds, covering a distance of 1 meter.
03

Calculate the velocity of the wave

The velocity of the wave is given by: \(v = \frac{Distance \: Travelled}{Time \: Taken} = \frac{1\: m}{2\: s}\) Hence, the velocity of the wave is 0.5 m/s. The correct answer is (C) \(0.5 \mathrm{~m} / \mathrm{s}\).

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Most popular questions from this chapter

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