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Simple Harmonic Motion Vidbook Pro SIMPLE HARMONIC MOTION Name Mech HW-79 Case A eum 1. Consider...

Simple Harmonic Motion

Vidbook Pro SIMPLE HARMONIC MOTION Name Mech HW-79 Case A eum 1. Consider the four block-and-spring(s) systems shown at rightMech Simple harmonic motion HW-80 The total potential energy of a system of multiple springs is defined to be the sum of the

Vidbook Pro SIMPLE HARMONIC MOTION Name Mech HW-79 Case A eum 1. Consider the four block-and-spring(s) systems shown at right. Each block moves on a horizontal, frictionless table. The blocks all have the same mass m, and all of the springs are identical and ideal, with spring constant k. At the instant shown, each block is released from rest a distance A to the right of its equilibrium position (indicated by the dashed line). In case B, assume that each spring is at its equilibrium length when the block is at its equilibrium position. a. Rank the cases according to magnitude of the net force on the block at the instant shown, from largest to smallest. (Hint: In case C, how far was the point connecting the two springs displaced when the block was displaced a distance A?) Explain. Case B Ulm Case C Hur U KA Case D Fute UUS b. Use your answers above to rank the cases according to the time it takes the block to return to its equilibrium position. Explain.
Mech Simple harmonic motion HW-80 The total potential energy of a system of multiple springs is defined to be the sum of the potential energies stored in each of the springs. Rank the cases according to total potential energy at the instant shown. (Hint: In each case, consider how much each individual spring is extended.) Explain. 2. Suppose that in cases B, C, and D. each combination of springs were replaced by a single spring such that the motion of the block is the same as before the replacement. constant of this new spring the effective spring constant of the original com Quantitatively, the effective spring constant, kar, may be defined by the relationship 2F block, spring = -keff X, where F block.spring is the sum of all forces on the block by the springs, and is the position of the block with respect to the equilibrium position. a. Rank the four cases according to kr, from largest to smallest. Explain. b. Use your ranking above and the relationship T = 2.7 m/ker to rank the cases according to period of oscillation. Explain. Is your ranking of the period consistent with your ranking in part b of problem 1? If not. resolve the inconsistency. c. Use your ranking from part a and the relationship Etotal = 1/2 keppA to rank the cases according to total energy. Explain. your ranking of the total energy consistent with your ranking in part c of problem 1? If not, resolve the inconsistency.
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Answer #1

Ans 1)

a) Magnitude of force in a spring = kx, where k is spring constant and x is displacement

for case A

net force = kA

for case B

net force = kA + kA = 2kA

keff = 2k

for case C

the two springs are in series so keff = (1/k + 1/k)-1 = k/2

therefore net force = kA/2

for case D

the two springs are parallel so keff = k +k = 2k

Net force = 2kA

Rank D=B>A>C

b)

more force that means more acceleration which means lesser time to travel.

Which means force and time are inversely related

Rank C>A>D=B

c)

Work done by each springs in each case

W = kxdc Jo

for case A

w = [kode = leva Jo

for case B

W = W spring1 + Wspring2 = kxdx + 1 kxdx = kA Jo

for case C

12 k(+2) W = W springi + Wspring2 = kxdr + 1 kudt = Jo

x1+x2 =A

(21 + 12) = 42

x+m2 = A? – 2.1172

:.: + 3 <A

this means potential energy of the two springs in case C is less than potential energy of the spring in Case A

for case D

W = W spring1 + Wspring2 = kxdx + 1 kxdx = kA Jo

Rank D=B>A>C

Ans 2

a)

effective spring constant is already found above for each case in Ans 1 a)

So, rank B=D>A>C

b)

Time period is inversely proportional to spring constant. So, more k means less T

So, rank C>A>D=B

c)

Ranking in terms of potential energy

we see the work done by each springs in the system

so potential energy = work done =1/2 kx2

For case A : potential energy of the spring = 1/2 kA2

For case B : potential energy = 1/2 *2k* A2 =-kA2

For case C : potential energy = 1/2 *k/2 * A2 = -1/4 kA2

For case D : potential energy = 1/2 * 2k* A2 =-kA2

Rank B=D>C>A

if any doubt feel free to comment

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