Question

# A smooth uniform string of natural length l, cross – sectional area A and Young’s modulus Y is pulled along its length by a force F on a horizontal surface. If the elastic potential energy stored in the string is $U = \dfrac{{{F^2}l}}{{xAY}}$, Find x.

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Hint: According to Mohr’s Law,
$Y = \dfrac{{Stress}}{{Strain}}$
Elastic Potential Energy is given by the formula,
$U = \int {dW}$

Complete step by step solution:
It is given in the question that,
Natural length of the string is l
Cross – sectional area of the string is A
Young’s Modulus of the string is Y
Force acting on the string is F

According to Mohr’s Law,
$Y = \dfrac{{Stress}}{{Strain}}$ ……….(1)
Where,
Y is the Young’s Modulus
Also,
$Stress = \dfrac{F}{A}$
Where,
F is the Tension in the String
And,
$Strain = \dfrac{{\Delta l}}{l}$
Inserting the values of stress and strain in equation 1,
We get,
$= > Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$
$= > Y = \dfrac{{Fl}}{{A\Delta l}}$
$= > F = \dfrac{{YA\Delta l}}{l}$ ……..(2)
Elastic Potential Energy is given by the formula,
$U = \int {dW}$
Where dW is the work done
In the above formula,
$dW = F.dx$
Where,
dx is the change in length of the string due to the applied tension
Inserting the value of dW in above equation,
We get,
$= > U = \int {F.dx}$
Inserting the value of F in the above equation and the limits would be from 0 to $\Delta l$ (Since, initially the string was unstretched so initial limit is 0 and finally the string is stretch by $\Delta l$ so the final limit is $\Delta l$)

We get,
$= > U = \int\limits_0^{\Delta l} {\dfrac{{YAx}}{l}.dx}$
$= > U = \dfrac{{YA}}{l}\int\limits_0^{\Delta l} {x.dx}$
$= > U = \dfrac{{YA}}{l}[\dfrac{{{x^2}}}{2}]_0^{\Delta l}$
$= > U = \dfrac{{YA}}{l}\dfrac{{{{(\Delta l)}^2}}}{2}$ …………(3)
From Equation 2,
$= > F = \dfrac{{YA\Delta l}}{l}$
$= > \Delta l = \dfrac{{Fl}}{{AY}}$
Inserting the value of $\Delta l$ in Equation 3,
We get,
$= > U = \dfrac{{YA}}{l}\dfrac{{{{(\dfrac{{Fl}}{{AY}})}^2}}}{2}$
$= > U = \dfrac{{{F^2}l}}{{2AY}}$
Comparing the above equation with the one given in the question,
$= > U = \dfrac{{{F^2}l}}{{2AY}} = \dfrac{{{F^2}l}}{{xAY}}$
Clearly, we can see from the above expression that,
$x = 2$

Note: Such types of questions are categorized into tricky sections. One requires in-depth knowledge of mechanics and calculus in order to solve such Calculation Intensive Problems.