Showing equivalent potential expressions for a Transverse String

[Potatochip911]

Homework Statement

I'm going through the derivation here starting on page 16. This image adds some context:

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Generalizing their result to the i'th particle they find the extended distance between two masses being ##\Delta l= \frac{(y_i-y_{i-1})^2}{2a}## Then since the potential energy is given by force * displacement they obtain $$V_i=\frac{F}{2a}(y_i-y_{i-1})^2$$ where ##F## is the tension in the string.

However since the string is stretching Hooke's law should apply which then results in the potential being ##V_i=\frac{1}{2}kx^2##, I'm having trouble showing this produces the same result as above

Homework Equations

##F=-kx##, where x is the extended distance

The Attempt at a Solution

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Ignoring the sign convention and using ##F=kx## we sub into the potential equation and get ##V_i=\frac{1}{2}Fx##, but x is just the extended distance ##\Delta l## therefore $$V_i = \frac{F}{2}\frac{(y_i-y_{i-1})^2}{2a}=\frac{F}{4a}(y_i-y_{i-1})^2$$ so I've obtained an extra factor of 1/2 but I can't seem to figure out why

 

[kuruman]

I think you missed the importance of the following statement in the derivation

It is also supposed that the tension in the string is F and that the displacements are sufficiently small that this is constant.

The effective spring constant is ##k = F/a## and the potential energy of the taut string is zero at equilibrium. When you displace the first mass straight up by (a very small) ##y_1##, its potential energy is ##V_1 = \frac{1}{2}(F/a)y_1^2##. When you displace the second mass straight up by ## y_2##, its potential energy is ##V_2 = \frac{1}{2}(F/a)(y_2-y_1)^2## and so on.

 

[Potatochip911]

I think you missed the importance of the following statement in the derivation

The effective spring constant is ##k = F/a## and the potential energy of the taut string is zero at equilibrium. When you displace the first mass straight up by (a very small) ##y_1##, its potential energy is ##V_1 = \frac{1}{2}(F/a)y_1^2##. When you displace the second mass straight up by ## y_2##, its potential energy is ##V_2 = \frac{1}{2}(F/a)(y_2-y_1)^2## and so on.

Hmmm, I can see that this gives the same result as the answer but I don't understand it. It's written in the form ##V_1=\frac{1}{2}(k)d^2## where you subbed in the effective spring constant however I can't understand why we are allowed to write the displacement of the string as merely the y-component when displacing along the y-axis results in a triangle being formed.