Question

Two ideal springs with spring constant k and relaxed length L are attached to two rigid walls, as shown in Figure 1. The two walls are a total distance D = 3L apart from one another. The springs are then attached to a ball that has mass m and negligible width, and the ball is then displaced from equilibrium, as shown in Figure 2. The location of the ball is given by the x-coordinate, as measured rightward from the left-most wall.

(a) Derive an appropriate
expression for U(x) , the
potential energy of the ball-double spring system expressed as a
function of c. *Express your answer in terms of m ,
k , l , and x .*

(b) What is the location of the
ball when the ball-double spring system is in equilibrium? What is
the potential energy of the system in that configuration?
*Express your answer in terms of m , k , and L
.*

*N**ote**: The symmetry of
this problem makes it is easy enough to guess the answer. In order
to get full credit,*

*ho**w**e**ver, you must verify your answer by
means of appropriate mathematical/physical reasoning.*

(c) The plot below shows U(x) for a ball-double spring system like the one described above. Using your answers from part (b), determine the numerical values of k and L for the potential energy curve U(x) plotted below.

(d) Suppose that the ball has mass m = 1.5 kg. If the ball is released from rest at 0.5 m, what is the speed of the ball when it reaches 2 m?

*F**OR FULL
CREDIT**: Show all relevant work and
commentary.* All numerical values must
include appropriate units, not just the final answer. Use extra
paper if necessary. Staple all of your work together for
submission. (*When in doubt, write it out.)*

Answer #1

a.)

If L<x<2L

Net expansion in the first spring = (x-L)

Net expansion in the second spring = (3L-x-L) = (2L-x)

If 0<x<L

Net compression in the first spring = (L-x)

Net expansion in the second spring = (3L-x-L) = (2L-x)

If x>2L

Net expansion in the first spring = (x-L)

Net compression in the second spring = (L- 3L-x) = ( x-2L )

U(x) =
0.5k(x_{1})^{2}+0.5k(x_{2})^{2}

= 0.5k [
x_{1}^{2}+x_{2}^{2}]

[For all the three cases U(x) is same as the function is
quadratic in x_{1} or x_{2}]

= 0.5k [ (x-L)^{2}+( x-2L)^{2}]

= 0.5k [ (x-L)^{2}+( x-2L)^{2} ]

= 0.5k [2x^{2} +5L^{2} - 6Lx ]

**U(x) = 0.5k [2x ^{2} - 6Lx +5L^{2}
] **

b.)

At equlibrium U(x) is minimum

Hence **0.5k [2x ^{2} - 6Lx +5L^{2}
] ** is min for

Hence **x = 1.5L**

U(1.5L) = 0.5k [ (1.5L-L)^{2}+( 1.5L-2L)^{2}] =
k [ (0.5L)^{2}] = **kL ^{2}/4** =

c.)

equlibrium position x = 1.5L = 1.5

**L= 1m**

U(1.5L) = 0.25kL^{2} = 1

**k = 4 Jm ^{2}**

d.)

Energy(x=0.5) = U(0.5) = 0.5k [2*0.5^{2} - 6L(0.5)
+5L^{2} ] = 2 [1/2- 3 +5 ] = 5

Energy(x=2) = U(2) + K(2) = 0.5k [2*2^{2} - 6L(2)
+5L^{2} ] + mv^{2} = 2 [8- 12 +5 ] +
1.5v^{2} = 2 + 1.5v^{2}

**Energy(x=0.5) = Energy(x=2)**

5 = 2 + 1.5v^{2}

1.5v^{2} = 3

v^{2} = 2

**v = sqrt(2) = 1.414 m/s**

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