Simple Harmonic Motion: Two Springs One Mass

In summary, the equation for the spring force in this system is F(spring system) = -kx - 3kx, and the solution for x(t) is x(t) = 0. This means that the mass will not move after the spring constant is changed.
  • #1
TheRedDragon
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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 attached to a mass m and two walls.

At a given instant, the right spring constant is somehow magically changed to 3k (the relaxed length remains L). what is the resulting x(t)?

Take the initial Position to be x=0

Homework Equations


The Attempt at a Solution



My problem is figuring out the Force equations of the system.

Could it be:
F(spring system) = -kx-3kx
where the 2L lengths on each side of the mass doesn't matter

or
F(spring system) = -k(2L+x)-3k(2L-x)
where the 2L lengths on each side of the mass does matter

or

Something else that I'm not considering.

I'm confused because the textbook doesn't have example problems on a similar system of two springs one mass. Also, my second guess would make a really complicated differential equation, where I don't think I could use use a guess such x(t)=Ae^(alpha*t) because there is a constant in the equation.

Can someone explain to me the correct equation for the spring force?
 
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  • #2


The correct equation for the spring force in this system would be F(spring system) = -kx - 3kx, where the 2L lengths on each side of the mass do not matter. This is because the change in the spring constant only affects the right spring, so the force from the left spring remains the same.

To solve for x(t), you can use the equation F = ma, where F is the total force from the two springs, m is the mass, and a is the acceleration. Since the mass is attached to the two springs, the acceleration will be the same for both springs.

So, F = ma becomes -kx - 3kx = ma. Rearranging this, we get a = -(4k/m)x. This is a second-order differential equation, so we can use the guess x(t) = Ae^(alpha*t) to solve it.

Substituting this into the equation, we get -(4k/m)Ae^(alpha*t) = -kAe^(alpha*t) - 3kAe^(alpha*t). Simplifying, we get alpha = ±√(4k/m), which gives us two possible solutions for x(t): x(t) = Ae^(-√(4k/m)t) and x(t) = Be^(√(4k/m)t).

To determine the values of A and B, we can use the initial conditions given in the problem. Since the initial position is x = 0, we can set x(t=0) = 0. This gives us the equation 0 = A + B.

To solve for A and B, we can use the initial velocity, which is given as 0. So, we can set v(t=0) = 0, which gives us the equation v(t=0) = A√(4k/m) - B√(4k/m) = 0. Solving for A and B, we get A = B = 0.

Therefore, the solution for x(t) is x(t) = 0, which means that the mass will not move after the spring constant is changed. This makes sense because the new spring constant (3k) is stronger than the original spring constant (k), so the mass will remain in its initial position.
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a system oscillates back and forth around a central equilibrium point with a constant amplitude and a fixed period of time.

2. How do two springs and one mass relate to simple harmonic motion?

In a system with two springs and one mass, the mass is attached to both springs and can move freely in one dimension. This creates a restoring force that causes the mass to oscillate back and forth, resulting in simple harmonic motion.

3. What factors affect the period of a simple harmonic motion in a system with two springs and one mass?

The period of a simple harmonic motion in this system is affected by the mass of the object, the spring constants of the two springs, and the initial displacement of the mass from its equilibrium position.

4. How is the amplitude of a simple harmonic motion determined in a system with two springs and one mass?

The amplitude of a simple harmonic motion in this system is determined by the initial displacement of the mass from its equilibrium position. The greater the initial displacement, the greater the amplitude of the oscillations.

5. Can the period of simple harmonic motion be changed in a system with two springs and one mass?

Yes, the period of simple harmonic motion in this system can be changed by altering the factors that affect it, such as the mass, spring constants, or initial displacement. For example, increasing the mass or spring constants will result in a longer period, while increasing the initial displacement will result in a shorter period.

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