# Showing equivalent potential expressions for a Transverse String

1. Apr 20, 2017

### Potatochip911

1. The problem statement, all variables and given/known data
I'm going through the derivation here starting on page 16. This image adds some context: .

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

2. Relevant equations
$F=-kx$, where x is the extended distance

3. The attempt at a solution

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

2. Apr 20, 2017

### kuruman

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.

3. Apr 20, 2017

### Potatochip911

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.