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Equation of a transverse simple harmonic progressive wave, travelling on a string is given by $y=\frac{\sqrt8}{π}cos(πx+πt)$. Find the ratio of maximum tension to minimum tension in the string.

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  • $\begingroup$ Hint: at some point in time, part of the string is at max. tension when the wave passes by. At a different point in time, that same location is at rest and doesn't "know" there's a wave pulse anywhere on the string. $\endgroup$ Apr 25, 2016 at 14:24
  • $\begingroup$ this question is screwing my nuts! if you get answer kindly ping me at chat.stackexchange.com/rooms/71/the-h-bar $\endgroup$ Apr 25, 2016 at 17:39
  • $\begingroup$ Same with me. Someone please help. $\endgroup$ Apr 26, 2016 at 4:36
  • $\begingroup$ i got answer finally!! but let me attach a photo because it involves lot of maths $\endgroup$ Apr 26, 2016 at 16:03
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    $\begingroup$ @DeNiSkA: please don't attach a photo. Either write out the equations or don't answer. $\endgroup$ Apr 27, 2016 at 6:20

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When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it.

Wave on string

If we take a segment small enough that we can consider it as a straight line, then the oscillation has stretched it from its rest length of $ds$ to the length $ds$. So if the rest tension is $T$ then the tension in the stretched segment is:

$$ T' = T \frac{ds}{dx} $$

So calculating the tension reduces to finding the ratio $ds/dx$. This can be related to the angle $\theta$ by:

$$ \frac{dx}{ds} = \cos\theta $$

and the angle $\theta$ is related to the gradient by:

$$ \frac{dy}{dx} = \tan\theta $$

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  • $\begingroup$ this is probably the best answer to this question. this comment is to enhance the flavor of this answer: "we assume the horizontal component of the tension in the string is constant" here we can even be sure that the the horizontal component of tension is equal because when we see from ground frame, center of mass moves only up and down but not right or left hence horizontal component of tension are equal $\endgroup$ Apr 27, 2016 at 11:40
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    $\begingroup$ What exactly is "The tension when the string is at rest" ? $\endgroup$ Apr 27, 2016 at 19:23
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    $\begingroup$ @BelalAhmed: This means the string is taut and it has not been disturbed to produce wave. $\endgroup$
    – user36790
    Apr 29, 2016 at 15:47
  • $\begingroup$ @John Rennie, I think in the second paragraph fourth line, its rest length should be $dx$, right? $\endgroup$
    – SarGe
    May 26, 2020 at 13:18
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The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations.

The tension is minimum and a constant, when no wave propagates through it. i.e., the velocity of the wave is zero. Now when a wave propagates through the medium, the tension increases and depends on the velocity (actually the frequency) of the wave. Find the tension at zero velocity and the velocity provided in the eqn. of the wave in your question. Find their ratio which will give you the required answer.

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    $\begingroup$ Typically we suggest posting hints as comments, not answers. $\endgroup$ Apr 25, 2016 at 14:22
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Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc

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    $\begingroup$ The velocity of the progressive wave is constant and v represents that. But the different points on the string at different instants will have different velocities but that will be particle's velocity and not the wave. $\endgroup$ Apr 25, 2016 at 13:36

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