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  • For the harmonic travelling wave y = 2 cos 2 π(10t-0.0080x+3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of a) 4 m

For the harmonic travelling wave y = 2 \cos 2 \pi (10t-0.0080x+3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of

a) 4 m

b) 0.5 m

c) \frac{\lambda }{2}

d) \frac{3\lambda }{4}

e) what is the phase difference between the oscillation of a particle located at x = 100 cm at t = Ts and t = 5s?

 

Answers (1)

y = 2 \cos 2 \pi (10t-0.0080x+3.5)

y=2\ cos\left (2\pi 10t-0.016\pi x+7\pi \right )

Wave is propagated in  +x direction because \omegat and kx are in opposite sign in 

standard equation y=a\cos\omega t-kx+\phi

a=2, \omega =20\pi , k=0.016\pi and \phi =7\pi

 apath difference=4m=400cm 

 phase difference \Delta \phi =\frac{2\pi }{\lambda } \times p=\frac{2\pi}{ \lambda} \times 400=k\times 400=0.016\pi \times 400 = 0.64 \pi rad

(b) path difference=0.5 m

\Delta \phi =\frac{2\pi}{\lambda} \times p=k\times p=0.016\pi \times 50=0.8 \pi rad

(c ) path difference=\lambda _{2}

\Delta \phi =\frac{2\pi }{\lambda} \times p=\frac{2\pi}{ \lambda} \times \frac{\lambda}{2} =\pi radian

d) \Delta \phi =\frac{2\pi}{\lambda} \times p=\frac{2\pi }{\lambda} \times \frac{3\lambda}{4} =\frac{3}{2}\pi radian

(e) T = \frac{2\pi}{\omega} =\frac{2\pi}{20\pi } =\frac{1}{10}sec

At x=100 , t=T

\phi _{1}=20\pi T-0.016\pi( 100)+7\pi =20\pi \times\frac{ 1}{10}-1.67\pi +7\pi =7.4\pi

At t=5s

\phi _{2}=20\pi (5)-0.016\pi 100+7\pi =100\pi -1.67\pi +7\pi =105.4\pi

\phi _{2}-\phi _{1}=98\pi radian

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infoexpert24

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