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The following equation represents standing wave set up in a medium , `y = 4 cos (pi x)/(3) sin 40 pi t` where `x and y` are in cm and t in second. Find out the amplitude and the velocity of the two component waves and calculate the distance adjacent nodes . What is the velocity of a medium particle at ` x = 3 cm` at time `1//8 s`? |
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Answer» From the knowledge of theory, if equation of statinary wave is `y=2Acoskxsinomegat,` then the component waves are `y_(1)=Asin(omegat-kx)` and `y_(2)=Asin(omegat+kx)` `:.` Amplitude of component waves `=A=(2A)/(2)=(4)/(2)=2cm` velocity of component waves `v=(omega)/(k)=(40pi)/(pi//3)=120cm//s=1.2m//s` (b) Wavelength, `lambda=(2pi)/(k)=(2pi)/(pi//3)=6cm.` Distance between adjacent nodes `=(lambda)/(2)=(6)/(2)=3cm` (c) Particle velocity `upsilon=(dy)/(dt)=(d)/(dt)[4cos((pix)/(3))sin40pit]` `160picos((pix)/(3))cos40pit` For `x=3cm ` and `t=(1)/(8)s` `upsilon=160picos((pixx3)/(3))cos40pixx(1)/(8)` `=160pi(-1)(-1)=160picm//s` `=1.6pim//s` |
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