Solving Wave Propagation Dilemma: Length Increase of 10%

AI Thread Summary
A wave traveling along a string takes 4 seconds to cover its length, which is then increased by 10%. The tension in the string is assumed to increase proportionally with the stretch, leading to a new wave speed that is also 10% greater. However, there is confusion regarding how the mass per unit length changes with the increase in length, as it decreases, potentially affecting the wave speed. Ultimately, the relationship between distance, speed, and time suggests that the time for the wave to travel the new length will also increase. The discussion highlights the complexities of wave propagation in relation to changes in string length and tension.
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Homework Statement



A wave takes 4s to travel form one end of the string to the other. Then the length is increased by 10%. Now how long does a wave take to travel the length of the spring?

Homework Equations



v = sqrt ( FL/m)
F = -kx

The Attempt at a Solution



If the original stretch was x, then an increase of ten percent would be:
1.1x.

So, if F = -kx, and since k is constant, if x increases by a factor of 1.1, then as does F. SO the tensional force is 1.1F the original.

v = sqrt ( 1.1F*L/m)
(delta d) / (delta t) = sqrt ( 1.1F*L/m)
taking the inverse:
(delta t) / (delta d) = sqrt ( m/1.1F*L)

where (delta d = 1.1L) since it has increased by ten percent.

So,

delta t = 1.1L*sqrt ( m/1.1F*L)
so the original delta t must increase by a factor of 1.1/sqrt(1.1)

?

?
 
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So, if F = -kx, and since k is constant, if x increases by a factor of 1.1, then as does F.
I don't understand this. The question doesn't say anything about the force or tension changing. Making the string longer might DEcrease the force but it sure wouldn't increase it.
 
It's actually a spring, but I was under the impression if you stretched a spring, you increased the tension..
 
Oh, sorry - I was thinking of putting a longer string on a guitar!

Using "v = sqrt ( FL/m)", it would appear that F increases by 1.1 and L increases by 1.1.
So the new v is 1.1 times the old v.
 
I was under the impression that L/m was a constant, regardless of whether it was stretched, since m/L is the linear mass density.
 
so L is the length? If so, L increases by a factor of 1.1, doesn't it?
The mass per unit length would decrease when the length increases.
 
I came here because apparently the velocity does not change.. I don't get how it doesn't change, though/
 
Oh, that's the answer I got! Using L increases by a factor of 1.1 and v increases by a factor of 1.1 in the formula t = d/v.
 
lol ill check it out thanks
 
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