- #1

Lito

- 9

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## Homework Statement

Two springs each have spring constant k and equilibrium length ℓ. They are both stretched a distance ℓ and then attached to a mass m and two walls (which are 4 ℓ apart).

At a given instant, the right spring constant is somehow magically changed to 3k (the relaxed length remains ℓ).

The question is: what is the resulting motion of the mass?

Take the initial position to be x = 0.

a. Show that the radial frequency of the new system is $$ \omega = 2 \sqrt{\frac{k}{m}} $$

b. Determine the new equilibrium position (relative to the original one).

c. Introduce a coordinate y that measures the position away from the new equilibrium position. Formulate the initial conditions and determine the motion y(t).

d. Determine the motion x(t) with respect to the original equilibrium position.

## Homework Equations

$$ F(x)= -kx $$

$$ x(t)= Acos(\omega t+\phi) $$

$$ v(t)= −A \omega sin(\omega t +\phi) $$

$$ a= −A \omega 2 cos(\omega t + \phi) $$

## The Attempt at a Solution

a.

Drawing the forces:

$$ 3kℓ - kℓ = m \ddot{x} $$

$$ 2kℓ = m \ddot{x} $$

But from this equation I don't get the result given in the question.

Any hint will be great :)

b. The equilibrium position is where the forces on the mass is 0, meaning that both of the spring forces deleting each other, but I'm not sure how to determine it.

I tried to describe the forces according to the displacement for every spring:

Left spring: -k(ℓ+x)

Right spring: 3k(ℓ-x)

But I'm not sure how to use/solve the equation:

3k(ℓ-x) = k(ℓ+x)

2kℓ = 4kx

x= 0.5ℓ

I'm not sure if my way is correct.

c. Does the mass moves in the y direction?

d. Any hint will be great :)

Thanks a lot !