Question

The ends of two identical springs are connected. Their unstretched lengths \(\ell\) are negligibly small and each has spring constant \(k\). After being connected, both springs are stretched an amount \(L\) and their free ends are anchored at \(y=0\) and \(x=\pm L\) as shown(Intro1figure). The point where the springs are connected to each other is now pulled to the position \((x, y)\). Assume that \((x, y)\) lies in the first quadrant.

$MPE_sp2_0.jpg$

What is the potential energy of the thetwo-spring system after the point of connection has been moved to position \(\left(x, y\right)\) ? Keep in mind that the unstretched length of each spring \({\ell}\) is much less than \({L}\) and can be ignored (i.e., \(\ell \ll L\) ).

Find the force   \(\vec{F}\) on the junction point, the point where the two springs are attached to each other.

Express  \(\vec{F}\) as a vector in terms of the unit vectors  \(\hat{x}\) and \(\hat{y}\).

Express the potential in terms of \({k}, x, y\), and \({L}\).

Answer 1

Concepts and reason

The concept required to solve the problem is potential energy stored in the spring and force acting on the spring system is used here.

First, find out the length of each of the spring and then use that value to find out the potential energy of two spring system.

In the second part, find out the compression in the spring and substitute that value in the expression of force.

Fundamentals

The expression of the potential energy stored in the spring system is given as follows:

$U = \frac{1}{2}k{x^2}$

Here, k is the spring constant and x is the spring compression.

The expression of the force acting on the spring system is given as follows:

$F = - kx$

(A)

The following figure shows the compression in the spring length.

The length of each of the spring is calculated as follows:

$\begin{array}{l}\\{L_1} = \sqrt {{{\left( {L - x} \right)}^2} + {y^2}} \\\\{L_2} = \sqrt {{{\left( {L + x} \right)}^2} + {y^2}} \\\end{array}$

The potential energy of the spring system is calculated as:

$U = \frac{1}{2}k{L_1}^2 + \frac{1}{2}k{L_2}^2$

Substitute $\sqrt {{{\left( {L - x} \right)}^2} + {y^2}}$ for ${L_1}$ and $\sqrt {{{\left( {L + x} \right)}^2} + {y^2}}$ for ${L_2}$ Lin the above expression.

$\begin{array}{c}\\U = \frac{1}{2}k{\left( {\sqrt {{{\left( {L - x} \right)}^2} + {y^2}} } \right)^2} + \frac{1}{2}k{\left( {\sqrt {{{\left( {L + x} \right)}^2} + {y^2}} } \right)^2}\\\\ = \frac{1}{2}k\left( {{L^2} + {x^2} - 2Lx + {y^2}} \right) + \frac{1}{2}\left( {{L^2} + {x^2} + 2Lx + {y^2}} \right)\\\\ = \frac{1}{2}k\left( {2{L^2} + 2{x^2} + 2{y^2}} \right)\\\\ = k\left( {{L^2} + {x^2} + {y^2}} \right)\\\end{array}$

(B)

The forces act on the spring are shown in the following figure:

The x and y components of the forces are given as follows:

$\begin{array}{l}\\{F_{1x}} = - k\left( {L + x} \right)\\\\{F_{1y}} = - ky\\\\{F_{2x}} = k\left( {L - x} \right)\\\\{F_{2y}} = - ky\\\end{array}$

The net force acting on the system is calculated as:

${F_{net}} = {F_{1x}} + {F_{1y}} + {F_{2x}} + {F_{2y}}$

Substitute $- k\left( {L + x} \right)$ for ${F_{1x}}$ , $- ky$ for ${F_{1y}}$ , $k\left( {L - x} \right)$ for ${F_{2x}}$ and $- kx$ for ${F_{2y}}$ the above expression.

$\begin{array}{c}\\{{\vec F}_{net}} = \left( { - k\left( {L + x} \right)} \right)\hat i + \left( { - ky} \right)\hat j + \left( {k\left( {L - x} \right)} \right)\hat i + \left( { - ky} \right)\hat j\\\\ = - 2kx\hat i - 2ky\hat j\\\end{array}$

Part B

[Part B]

Ans: Part A

The potential energy of the spring system is $k\left( {{L^2} + {x^2} + {y^2}} \right)$ .

Part B

The net force acting on the junction is $- 2kx\hat i - 2ky\hat j$ .

Answer 2

length of the left spring = R1 = √(y² +(L+x)²)
length of the right spring = R2 = √(y² +(L-x)²)
Total PE = (1/2)k (R1² + R2²)
= k (x² + y² + L²)
The x- and y- components of the forces are
F1_x = -k R1 cos θ (where θ is the anglebetween x-axis &spring) = -k (L+x)
F1_y = -k R1 sin θ = -k y
F2_x = -k R2 cos α (where α is the anglebetween x-axis &spring) = -k (L+x)
F1_y = -k R2 sin α = -k y
The net force F is the vector sum of F1 and F2.
F = -2kx i - 2ky j
where i and j are the 2 unit vectors.

hope this helps!