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  • A travelling harmonic wave on a string is described by what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s ? Is this velocity equal to the velocity of wave propagation?

Q.15.22 (a) A travelling harmonic wave on a string is described by

 y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)
(a) what are the displacement and velocity of oscillation of a point at x = 1 \: cm, and t = 1\: s ? Is this velocity equal to the velocity of wave propagation?

 

Answers (1)

best_answer

y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)

The displacement of oscillation of a point at x = 1 cm and t = 1 s is

\\y(1,1)=7.5sin(0.0050\times 1+12\times 1+\frac{\pi }{4})\\ y(1,1)=7.5sin(12.79)\\ y(1,1)=7.5sin(\frac{12.266\times180^{o} }{\pi })\\ y(1,1)=7.5sin(733.18^{o})\\ y(1,1)=1.71cm

The general expression for the velocity of oscillation is

\\v_{y}(x,t)=\frac{\mathrm{dy(x,t)} }{\mathrm{d} t}\\=\frac{d}{dt}\left [ sin\left ( 7.5sin(0.0050x+12t+\frac{\pi }{4} \right ) \right ]\\ =90cos(0.0050x+12t+\frac{\pi }{4})

\\v_{y}(1,1)=90cos(0.0050\times 1+121+\frac{\pi }{4})\\ =90cos(12.79)\\ =90cos(733.18^{o})\\ =87.63\ cm\ s^{-1}

y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)

k=0.005 cm-1

\omega =12rad/s

The velocity of propagation of the wave is

\\v=\nu \lambda \\ v=\frac{\omega }{2\pi }\times \frac{2\pi }{k}\\ v=\frac{\omega }{k}\\ v=\frac{12}{0.005\times 100}\\ v=24ms^{-1}

The velocity of oscillation of point at x = 1 cm and t = 1 cm is not equal to the propagation of the wave.

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Sayak

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