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In the given progressive waves y = 5 \sin (100 \pi t -0.4 \pi x) where y and x are in m, t is in s. What is the

a) amplitude

b) wavelength

c) frequency

d) wave velocity

e) particle velocity amplitude

 

Answers (1)

Given the progressive wave equation

y=a\sin \omega t-kx+\phi   

y = 5 \sin (100 \pi t -0.4 \pi x) 

(a) Amplitude a=5m

( b )k=\frac{2\pi }{\lambda}=0.4\pi 

Wavelength

\lambda =\frac{2\pi}{k} =\frac{2\pi }{0.4\pi }=5m

(c) \omega =2\pi v;v=\frac{\omega}{2\pi} , where \omega =100\pi

v=\frac{100\pi}{2\pi} =50Hz

 Wave velocity

V=v\lambda =50\times 5=\frac{250m}{s}

(e) Particle velocity in the direction of amplitude at a distance x from source

 y=5\sin 100\pi t-0.4\pi x

 \frac{dy}{dt}=5\times 100\pi \cos\left (100\pi t-0.4\pi x \right )

 Maximum velocity of the particle is at its mean position

 \cos100 \pi t-0.4\pi x=1

100\pi t-0.4\pi x=0

\frac{dy}{dt}=5\times 100\pi =500 \pi m/s

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