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Chapter 10: Problem 1496

A wave travelling along a string is described by the equation \(\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx}) .\) The maximum particle velocity is \(\ldots \ldots\) (A) A\omega (B) \(\omega / \mathrm{k}\) (C) \(\mathrm{d} \omega / \mathrm{dk}\) (D) \(\mathrm{x}\)

Short Answer

Expert verified
The maximum particle velocity can be found by deriving the wave equation with respect to time: \(\frac{dy}{dt} = A \cos(\omega t - kx) \cdot \omega\). The maximum value occurs when \(\cos(\omega t - kx) = 1\), resulting in a maximum particle velocity of \(A\omega\), which corresponds to option (A).

Step by step solution

01

Derive the wave equation with respect to time

We will derive the wave equation, \(y=A \sin(\omega t-kx)\), with respect to time (t) to get the particle velocity, as the velocity represents the rate of change in vertical displacement (y) with respect to time: \[\frac{dy}{dt} = A \frac{d}{dt} [\sin(\omega t - kx)]\]
02

Apply the chain rule

Here, we will apply the chain rule because we have a composition of functions, i.e., a function of \(\omega t - kx\): \[\frac{d}{dt} [\sin(\omega t - kx)] = \cos (\omega t - kx) \cdot \frac{d}{dt} [\omega t - kx]\]
03

Derive \(\omega t - kx\) with respect to time

Derive the inner function \((\omega t - kx)\) with respect to time (t): \[\frac{d}{dt} [\omega t - kx] = \omega\]
04

Multiply to get the particle velocity

Multiply the results from step 2 and step 3 to get the particle velocity: \[\frac{dy}{dt} = A \cos(\omega t - kx) \cdot \omega\]
05

Find the maximum particle velocity

To find the maximum particle velocity, we need to find the maximum value of \(\frac{dy}{dt}\). We know that the maximum value of \(\cos(\omega t - kx)\) is 1. Therefore, the maximum particle velocity will be when \(\cos(\omega t - kx) = 1\): \[\frac{dy}{dt} = A \cdot 1 \cdot \omega\] Therefore, the maximum particle velocity is \(A\omega\), which corresponds with option (A).

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