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Two travelling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a standing wave having the equation y = A cos kx sin cot in which A = PO mm, k = P57 cm and co = 78'5 s (a) Find the velocity of the component travelling waves. (b) Find the node closest to the origin in the region x> 0. (c) Find the antinode closest to the origin in the region x> 0. (d) Find the amplitude of the particle at x 2.33 cm.

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2 years ago Saurabh Koranglekar
8406 Points
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6 months ago
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6 months ago
```							Here, the equation of standing wave is  y=Acoskxsinωty=Acoskxsinωtwhere  A=1.0mm=0.1cm,k=1.57cm,ω=78.5sAs travelling waves are of equal amplitudes equal frequencies and moving in opposite directions, their equations can be written as  y1=1=  A/2  sin(ωt−kx)y22=A/2  sin(ωt+kx)y=y1+y2 =A/2  [sin(ωt−kx)+sin(ωt+kx)]=Acoskxsinωt.(a) Wave velocity of either waveυ=ωk=78.5s/1.57c=50cm/sυ=ωk=78.5s-11.57cm-1=50cm/s(b) For a node,  y=0,∴coskx=0,kx=π/2x=π/2 whole divided by  k=3.14/2 whole divided by  1.57=1cm(c) For an antinode,  y=max,for which  |coskx|=1kx=nπ=π  for smallest value of x  x=π/k=3.14/1.57=2cm(d) Teh amplitude of vibration  =|Acoskx|r=1.0cos(1.57×2.33radian)=1.0cos(3.658rad)=1.0cos(3.658×57degree)=1.0cos(209∘)= -cos29∘r=−0.875mm
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6 months ago
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