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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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Most popular questions from this chapter

If the equation of a wave in a string having linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}\( is given by \)\mathrm{y}=0.02\( \)\sin [2 \pi\\{1 /(0.04)\\}-\\{\mathrm{x} /(0.50)\\}]$, then the tension in the string is \(\ldots \ldots \ldots \ldots\) N. (All values are in \(\mathrm{mks}\) ) (A) \(6.25\) (B) \(4.0\) (C) \(12.5\) (D) \(0.5\)

As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and $\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}$ and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).What will be the periodic time of the block, between the two springs? (A) \(1+(5 \pi / 6) \mathrm{s}\) (B) \(1+(7 \pi / 6) \mathrm{s}\) (C) \(1+(5 \pi / 12) \mathrm{s}\) (D) \(1+(7 \pi / 12) \mathrm{s}\)

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

A metal wire having linear mass density \(10 \mathrm{~g} / \mathrm{m}\) is passed over two supports separated by a distance of \(1 \mathrm{~m}\). The wire is kept in tension by suspending a \(10 \mathrm{~kg}\) mass. The mid point of the wire passes through a magnetic field provided by magnets and an a. c. supply having frequency \(\mathrm{n}\) is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a. c. supply? (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

When a mass \(M\) is suspended from the free end of a spring, its periodic time is found to be \(\mathrm{T}\). Now, if the spring is divided into two equal parts and the same mass \(\mathrm{M}\) is suspended and oscillated, the periodic time of oscillation is found to be \(\mathrm{T}\) '. Then \(\ldots \ldots \ldots\) (A) \(\mathrm{T}<\mathrm{T}^{\prime}\) (B) \(\mathrm{T}=\mathrm{T}^{\prime}\) (C) \(\mathrm{T}>\mathrm{T}^{\prime}\) (D) Nothing can be said.
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