A block of mass m is free to move in the horizontal direction on a frictionless

Question

A block of mass m is free to move in the horizontal direction on a frictionless surface. Two massless springs are attached in seires in between the block and an immovable wall. The springs have spring constants k1 and k2. The block is at equilibrium at x=0.

Now we displace the block to the right by distance delta(x). (Let us define the positive x axis to the right. Let i be the unit vector in the positive x direction.)

a. By how much is spring 1 extended?

b. By how much is spring 2 extended?

c. What is the magnitude and direction of the force spring 1 exerts on the wall?

d. What is the total force the block feels? It should be in the form F= -keff delta(x), where keff is the effective spring constant of the system of springs. What is keff? (Hint: keff should be positive. Think about the direction of the forces if you got keff to be negative.)

e. Given the effective spring constant and the mass of the block, what is the period of the simple harmonic motion the block undergoes?

f. Now consider the situation shown in the figure above, where the springs are in parallel. What is the effective spring constant keff now, and what is the period of simple harmonic motion the block undergoes?

I absolutely have no idea how to even tackle this, so an elaborate explanation will be much appreciated.

Explanation / Answer

When you have springs in series, the effective spring constant is given by 1/k = 1/k1 + 1/k2 And compressed distance is given by x1*k1 = x2*k2 a) Here we've displaced the block a distance i from its equilibrium, so that x1+x2 = i. Both springs will also be extended (or compressed) in the positive direction. And we'll have: x1*k1 = (i-x1)*k2 x1 =i*k2 / (k1+k2) b) In the same way, we'll have x2 = i*k1 / (k1+k2) c) Force in a spring follows Hooke's law F = -kx For spring 1, we have, unsing answer from a) F1 = -k1 x1 = -i*k1*k2 / (k1+k2) And for F2, we have F2 = -i*k1*k2 / (k1+k2) d) To find the total force, we can use k as defined at the top, so that F = -1 / (1/k1 + 1/k2) * i We also know it'll the sum of F1 and F2 F = -2i*k1*k2 / (k1+k2) A little algebra will convince you that bot expresions are exactly the same. e) The period for an oscillating spring is given by T = 2pv(m/k) = 2pv( m / (1/k1+1/k2) ) f) With springs in parallel, you can simply add the different constants together, so that k = k1 + k2. Here, T = 2pv(m/k) becomes T = 2pv(m/(k1+k2))

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