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Chapter 18: Problem 28

A string vibrates according to the equation $$ y=(0.520 \mathrm{~cm}) \sin [(1.14 \mathrm{rad} / \mathrm{cm}) x] \cos [(137 \mathrm{rad} / \mathrm{s}) t] $$ (a) What are the amplitude and speed of the component waves whose superposition can give rise to this vibration? (b) Find the distance between nodes. (c) What is the velocity of a particle of the string at the position \(x=1.47 \mathrm{~cm}\) at time \(t=\) \(1.36 \mathrm{~s} ?\)

Short Answer

Expert verified
The amplitude is 0.520cm and the speed is 0.00832 cm/s, the distance between nodes is 5.51 cm, and the velocity at x=1.47 cm and t=1.36 s is -96.83 cm/s.

Step by step solution

01

Identify the Amplitude and Speed

The amplitude of a wave is the maximum displacement from its equilibrium position, and is given by the coefficient of the equation. In this case, it is \(0.520cm\). The wave speed is given by \(\frac{wavenumber}{angular frequency}\), so in this case it is \(\frac{1.14 rad/cm}{137 rad/s} = 0.00832 cm/s\).
02

Find the Distance Between Nodes

A node is a point where the displacement of the wave is always zero. The distance between nodes is called the wavelength, and it is given by \(\frac{2\pi}{wavenumber}\). So in this case it is \(\frac{2\pi}{1.14 rad/cm} = 5.51 cm\).
03

Find the Velocity at Given Point

The velocity of a point in the wave is given by the derivative of the displacement function. Therefore the velocity is \(velocity = \frac{d}{dt}(0.520cm \sin(1.14 rad/cm * 1.47cm) * cos(137 rad/s * 1.36s))\), which simplifies to \(-96.83 cm/s\).

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