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haridas_mandal Group Title

The equation of a stationary wave in a string is Y = 4 mm sin [(3.14m^-1)x] cos wt. Select the correct alternatives:a) The amplitude of component waves is 2mm. b) The amplitude of component waves is 4mm. c) The smallest possible string is 0.5m. d) The smallest possible length of string is 1.0m 1) b,c 2) b,d 3) a,c 4) a,d Pl explain the answer

  • one year ago
  • one year ago

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  1. yrelhan4 Group Title
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    *

    • one year ago
  2. haridas_mandal Group Title
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    Can't see the reply

    • one year ago
  3. yrelhan4 Group Title
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    well i just bookmarked the question. i will know when somebody answers it.

    • one year ago
  4. experimentX Group Title
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    I hope the equation is \[ y(x,t) = 4 \sin ( \pi x) \cos (\omega t) \] Here is a demonstration of the wave If you have Mathematica, you can try this code ` Animate[Plot[ 4 Cos[Pi x] Cos[3 t], {x, -5, 5}, ` ` PlotRange -> {-5, 5}], {t, 1, 5}]` The amplitude is the maximum value of displacement from the mean position. As you can see \[ |4 \sin ( \pi x) \cos (\omega t)|\le 4 \cdot 1 \cdot 1 = 4 \] The amplitude is 4. For min length of the wire required, we need half wave length. For that .. keep t constant. Now we have $$ 4 \cos ( \pi x ) $$ whose period is 2.|dw:1359403049202:dw| Hence the min-required string is 1mm

    • one year ago
  5. haridas_mandal Group Title
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    Correct answers are already there in the option a-d. We need to get the right combination as is given in 1-4. Option b is correct as amplitude is 4mm. As you have explained smallest length of string is 1.0 m we get option 2) b,d as the right answer. I do not have mathematica and I have not followed you as to how you derived the smallest length to be 1 m . If you explain in some other method I think I will be able to understand and make my daughter understand.

    • one year ago
  6. experimentX Group Title
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    look at this demonstration http://demonstrations.wolfram.com/TransverseStandingWaves/ |dw:1359412263896:dw| Imagine a graph of your function. When you stop time, you will get ordinary cosine function. Try to find distance (between two points) of such shape in your graph.

    • one year ago
  7. haridas_mandal Group Title
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    "For min length of the wire required, we need half wave length. For that ..keep t const. Now we have 4 cos(#x) whose time period is 2." This part I am not understanding..and how it is ingetting interlinked to smallest length of the string.pl help.

    • one year ago
  8. experimentX Group Title
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    EDIT:"... Now we have \( 4 \cos(\pi x) \) whose time is 2." |dw:1359443751078:dw||dw:1359443786477:dw| The minimum possible length is the half the wavelength.

    • one year ago
  9. haridas_mandal Group Title
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    I think I am deficient in some basics. Pardon me , its long long time since I last prepared for my board exam in 1972.but i am determined.My basic doubts: i) How you are reducing the equation to 4 cos(pie x) ii) And deducing Time Period to 2 ? iii) Is this equn representing 4 Cos (kx ) or 4 cos (wt) where T =2pie/w=2pie/pie=2 I am literally confused. iv) For a a string rigidly tied at both ends it will have two nodes at the ends and an anti-node at the middle. How come your drawing while shifting one of the rigid body contradicting the same.. Pl help.

    • one year ago
  10. shubhamsrg Group Title
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    It is 4 sin(kx) cos(wt) k= 3.14 = 2pi/ l where l denotes wavelength => l =2 => min length = 2/2 =1 also amplitude = 4 clearly. As @experimentX rightly said.

    • one year ago
  11. shubhamsrg Group Title
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    You can also view it as superposition of 2 waves => 2sin(kx + wt) + 2sin(kx -wt)

    • one year ago
  12. shubhamsrg Group Title
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    both waves are identical just that there direction is opposite.

    • one year ago
  13. shubhamsrg Group Title
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    => 2sin(wt + kx) - 2sin(wt -kx)

    • one year ago
  14. experimentX Group Title
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    @haridas_mandal i think you are not understanding the basic notion of wave motion. y = A sin(kx) <-- this is your usual Sine graph |dw:1359471231793:dw| y = A sin(kx + wt) this is your progressive wave. |dw:1359471264055:dw| the difference is time is also flowing ... (kx+wt) <-- is also always changing. so this is always dynamic like animation http://en.wikipedia.org/wiki/File:Simple_harmonic_motion_animation.gif when two same waves ... one going toward the +ve direction and another going to -ve direction. The result is standing wave. http://upload.wikimedia.org/wikipedia/commons/7/7d/Standing_wave_2.gif |dw:1359473405983:dw|

    • one year ago
  15. experimentX Group Title
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    |dw:1359473577173:dw| If you look at the standing wave, you see that there are two things that are changing ... on is 'x' and other is 't'. when x changes ,,, as you go along, 'x', you note that you get wiggling sine curves. and this 't' animates. And special thing about standing wave is ... it does not move forward. so you are only looking for part static part ... which you get when you stop the animation. So what do you do? ... stop the time in that equation ... or remove it. or don't change it ... and find the period in 'x' (i.e. along the distance). which is what you get.

    • one year ago
  16. haridas_mandal Group Title
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    My daughter has seen it and well understood. I have some basic doubt but I shall leave it to that ( according to her 4 Sin kx ( 4 sin(pie x) )is the amplitude and is maxm when SinKx is 1 (one) or Amplitude is 4) . Now K = 2 pie /Lamda hence Lamda = 2 pie/k =2 pie / pie ( as here K= pie (given in the equation) , so lamda is equal to 2 m, for shortest length of String it is to be length = Lamda /2 i.e equal to 1 meter. So option 2 is correct. I hope her reasoning are correct. If not pl respond or else I must thank you especially experimentX for taking so much trouble .Meanwhile I need to refresh on this chapter.

    • one year ago
  17. experimentX Group Title
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    well ... period of Sin(x) = 2 Pi Period of Sin(kx) = 2Pi/k ... this should work. Basically if you multiply the x thing inside, ... that happens it contracts and ... the period decreases by ... 1/k. It should be okay!!

    • one year ago
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