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Chapter 10: Problem 1513

Standing waves are produced by the superposition of two waves \(\mathrm{y}_{1}=0.05 \sin (3 \pi \mathrm{t}-2 \mathrm{x})\) and \(\mathrm{y}_{2}=0.05 \sin (3 \pi \mathrm{t}+2 \mathrm{x})\) where \(\mathrm{x}\) and \(\mathrm{y}\) are in meters and \(\mathrm{t}\) is in seconds. The amplitude of a particle at \(\mathrm{x}=0.5 \mathrm{~m}\) is \(\ldots \ldots\) (A) \(1.08 \times 10^{-1} \mathrm{~m}\) (B) \(5.4 \times 10^{-2} \mathrm{~m}\) (C) \((\pi / 2) \times 10^{-1} \mathrm{~m}\) (D) \(\pi \times 10^{-1} \mathrm{~m}\)

Short Answer

Expert verified
The amplitude of a particle at x = 0.5m is approximately \((A) 1.08 \times 10^{-1} \mathrm{~m}\).

Step by step solution

01

Write down the given waves:

We are given two waves: \(y1 = 0.05sin(3\pi t - 2x)\) \(y2 = 0.05sin(3\pi t + 2x)\)
02

Calculate the displacement of the particle at x = 0.5m:

We need to find the value of y1 and y2 at x = 0.5m. For \(y1\), substituting x = 0.5m, we get: \(y1 = 0.05sin(3\pi t - 2(0.5)) = 0.05sin(3\pi t - 1)\) For \(y2\), substituting x = 0.5m, we get: \(y2 = 0.05sin(3\pi t + 2(0.5)) = 0.05sin(3\pi t + 1)\)
03

Apply the superposition principle:

The principle of superposition states that the resulting wave is the sum of the individual waves. Therefore, we will add y1 and y2: \(y = y1 + y2 = 0.05sin(3\pi t - 1) + 0.05sin(3\pi t + 1)\)
04

Find the amplitude:

The amplitude of the particle is the maximum displacement from its equilibrium position. Using the identity \(sin(A)+sin(B)=2sin(\frac{A+B}{2})cos(\frac{A-B}{2})\), we rewrite y: \(y = 2(0.05)sin(\frac{(3\pi t - 1) + (3\pi t + 1)}{2})cos(\frac{(3\pi t - 1) - (3\pi t + 1)}{2})\) Simplifying the expressions: \(y = 0.1sin(3\pi t)cos(-1)\)
05

Compare the amplitude to the given options

The amplitude is the coefficient of the sine term, which is 0.1 or \(1.0 \times 10^{-1}m\). However, this value does not match any of the given options. The closest option to this amplitude is (A) \(1.08 \times 10^{-1} \mathrm{~m}\).

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