# Harmonic motion- Two springs, One mass

• Lito
In summary: That is the new equilibrium position, and x is a perfectly good variable for measuring the position of the mass from that equilibrium position. If you want to make it clearer, you could say that x(t) is the displacement of the mass from the new equilibrium position. It's not a big deal, but your final equation would be more consistent with your preceding text if you said something like that.
Lito

## 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 !

In (a), you are asked about subsequent motion. The extension l is only the initial state. When the displacement is x, what will the tensions in the springs be? (My inclination would be to solve b first, then use that as the x=0 for part d, then deduce a.)
Your solution to b is correct.
I think you are only supposed to consider motion in the x direction.

Lito
haruspex said:
My inclination would be to solve b first, then use that as the x=0 for part d, then deduce a.

I tried to solve using the new equilibrium position combining with the equations:

The new equilibrium position is x= 0.5ℓ
Therefor the Amplitude is $$A=0.5ℓ$$
Solving
$$\dot{x}(0)= 0$$
$$\dot{x}(0)= 0.5ℓ \omega cos(\phi) = 0$$
$$\phi = 0$$

The forces in the x(0)
$$2kℓ = m \ddot{x}(0)$$
$$\ddot{x}(0) = 2kℓ/m$$
$$\ddot{x}(0) = a= − 0.5ℓ ω^2cos(ωt+ϕ)$$
$$2kℓ/m = − 0.5ℓ ω^2cos(0)$$
$$ω^2 = 4k/m$$
$$ω = 2 \sqrt{k/m}$$

Therefore
$$x(t)= 0.5ℓ cos(2 \sqrt{k/m}*t)$$

Does is make sense ?

A few slips in there. It's not clear whether you are taking x(t) as the sine or the cosine of ##\omega t+\phi##. You have both ##\dot x## and ##\ddot x## using the cosine. Also you took cos(0) to be 0. Somehow this led to a negative value for ##\omega^2##, but you wisely ignored that.
In terms of the problem as set, do you think your final equation is for x(t) or y(t)? Does it satisfy the initial conditions?

Lito
haruspex said:
A few slips in there. It's not clear whether you are taking x(t) as the sine or the cosine of ##\omega t+\phi##. You have both ##\dot x## and ##\ddot x## using the cosine. Also you took cos(0) to be 0. Somehow this led to a negative value for ##\omega^2##, but you wisely ignored that.
In terms of the problem as set, do you think your final equation is for x(t) or y(t)? Does it satisfy the initial conditions?

Hey, Thanks so much !

I noticed the mistake now, ##\dot x## with sine, it supposed to be:
##\dot x(0) = -0.5ℓωsin(ϕ)=0##
==> ## ϕ=0 ##

about the negative value i wasn't sure if the equation is correct, should I put negative signs for both sides?
The final equation is for x(t).

Lito said:
Hey, Thanks so much !

I noticed the mistake now, ##\dot x## with sine, it supposed to be:
##\dot x(0) = -0.5ℓωsin(ϕ)=0##
==> ## ϕ=0 ##

about the negative value i wasn't sure if the equation is correct, should I put negative signs for both sides?
The final equation is for x(t).
Looking at your final equation, where is the equilibrium? Where should it be for x?

Lito
haruspex said:
Looking at your final equation, where is the equilibrium? Where should it be for x?

Corresponding with the initial x, the new equilibrium position is x=0.5ℓ.

Lito said:
Corresponding with the initial x, the new equilibrium position is x=0.5ℓ.
Yes, but is that what your final equation in post #3 says? What, according to that equation, is the range of x?

haruspex said:
Yes, but is that what your final equation in post #3 says? What, according to that equation, is the range of x?
ohh i miss the displacement ? The spring should range from 0 to ℓ therefore:
$$x(t)= 0.5ℓ - 0.5ℓ cos(2 \sqrt{k/m}*t)$$

Lito said:
ohh i miss the displacement ? The spring should range from 0 to ℓ therefore:
$$x(t)= 0.5ℓ - 0.5ℓ cos(2 \sqrt{k/m}*t)$$
Yes.

Lito

## 1. What is harmonic motion?

Harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium. This results in the motion of the object being sinusoidal or wave-like in nature.

## 2. How are two springs involved in harmonic motion?

In the case of two springs and one mass system, the mass is connected to two springs in parallel. This means that the two springs are attached at either end of the mass and pull in opposite directions when the mass is displaced.

## 3. What is the role of the mass in harmonic motion?

The mass in harmonic motion serves as the object that is undergoing the periodic motion. It experiences a restoring force from the springs and is responsible for the oscillatory motion of the system.

## 4. How is the period of harmonic motion affected by the mass and springs?

The period of harmonic motion is affected by the mass and the stiffness of the springs. A heavier mass or stiffer springs will result in a longer period, while a lighter mass or less stiff springs will result in a shorter period.

## 5. What are the applications of harmonic motion with two springs and one mass?

Harmonic motion with two springs and one mass has many real-world applications, including in the design of suspension systems, musical instruments, and oscillators in electronics. It is also used in studying the behavior of waves and vibrations in various systems.

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