Question

Two ideal springs with spring constant k and relaxed length L are attached to two rigid...

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 .

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

however, 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?

FOR 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.)

Homework Answers

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(x1)2+0.5k(x2)2

= 0.5k [ x12+x22]

[For all the three cases U(x) is same as the function is quadratic in x1 or x2]

= 0.5k [ (x-L)2+( x-2L)2]  

= 0.5k [ (x-L)2+( x-2L)2 ]  

= 0.5k [2x2 +5L2 - 6Lx ]  

U(x) = 0.5k [2x2 - 6Lx +5L2 ]  

b.)

At equlibrium U(x) is minimum

Hence 0.5k [2x2 - 6Lx +5L2 ]   is min for x = 6L/(2*2) = 3L/2

Hence x = 1.5L

U(1.5L) = 0.5k [ (1.5L-L)2+( 1.5L-2L)2] = k [ (0.5L)2] = kL2/4 = 0.25kL2

c.)

equlibrium position x = 1.5L = 1.5

L= 1m

U(1.5L) = 0.25kL2 = 1

k = 4 Jm2

d.)

Energy(x=0.5) = U(0.5) = 0.5k [2*0.52 - 6L(0.5) +5L2 ] = 2 [1/2- 3 +5 ] = 5

Energy(x=2) = U(2) + K(2) = 0.5k [2*22 - 6L(2) +5L2 ] + mv2 = 2 [8- 12 +5 ] + 1.5v2 = 2 + 1.5v2

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

5 = 2 + 1.5v2

1.5v2 = 3

v2 = 2

v = sqrt(2) = 1.414 m/s

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