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Grade: 12

                        

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.

2 years ago

Answers : (3)

Saurabh Koranglekar
askIITians Faculty
8406 Points
							Dear student

Please write the Question in a standard form or attach an image of the question

Regards
6 months ago
Vikas TU
12280 Points
							
Dear student 
Question is not clear 
Please attach an image, 
We will happy to  help you 
Good Luck
Cheers
6 months ago
Arun
24734 Points
							

Here, the equation of standing wave is  
y=Acoskxsinωt
y=Acoskxsinωt
where  
A=1.0mm=0.1cm,k=1.57cm,
ω=78.5s
As 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)
y2
2=A/2  sin(ωt+kx)
y=y1+y2 =
A/2  [sin(ωt−kx)+sin(ωt+kx)]
=Acoskxsinωt.
(a) Wave velocity of either wave
υ=
ωk=
78.5s/1.57c
=50cm/s
υ=ωk=78.5s-11.57cm-1=50cm/s
(b) For a node,  
y=0,
∴coskx=0,kx=π/2
x=
π/2 whole divided by  k
=3.14/2 whole divided by  1.57
=1cm
(c) For an antinode,  
y=max,
for which  
|coskx|=1
kx=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
 
6 months ago
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