Solving a Mass-Spring System: Max. Displacement of Mass

In summary, the conversation is about a 1.2kg mass attached to a linear spring on a horizontal surface with kinetic friction. The speed of the mass as a function of position, maximum speed and position, distance the spring is compressed, and maximum displacement of the mass after each half oscillation are discussed. The equations of conservation of energy are used to solve for these values. The final part, deriving an expression for the maximum displacement of the mass after each half oscillation is not solved.
  • #1
dozappp
4
0

Homework Statement


A block of mass 1.2kg, resting on a horizontal surface of coefficient of kinetic friction of .15, is attached to a linear spring of force constant 12.0 N/m. When the spring is unstretched the mass is at a position x=0. THe mass is pulled to the right streching the spring a distance x=0.8 m, and relased from rest at t=0. The mass accelerates the the left and momentarily comes to rest at point B.

a) Derive an expression for the speed of the mass as a function of x.
b)Calculate the maximum speed of the mass during this time. At what position x does this maximum speed occur?
c)Calculate the distance the spring is compressed whne the mass stops momentarily at point B.
d)Derive an expression for the maximum displacement of the mass at the end of each half oscillation as a function of hte number of half oscillations, n.


Homework Equations


Conservation of energy is what i used for a, b, and c. No idea how to do d.


The Attempt at a Solution



a)
.5*k*.8^2-.5*k*x^2-m*g*.15*(.8-x)=.5*m*v^2
k=12, m=1.2, g =9.8
sqrt(2(6*.8^2-6*x^2-1.764(.8-x))))=v
which simplifies to
v=Sqrt[-10x^2+2.94x+4.04]

b)
v=Sqrt[-10x^2+2.94x+4.04]
dv/dx = (-Sqrt[10]*(x-.147))/Sqrt[-x^2+.294+.404]
dv/dx = 0
x=.147
V(.147) = 2.06 m/s
therefore
2.06 m/s @ x=.147

c)
V=0
Sqrt[-10x^2+2.94x+4.04] = 0
x = -.505

.505 m compressed

d)

OK, I think I got the previous parts correct, but this part seriously breaks my brain. I simply do not know where to start. I tried to make a function like

1/2*k*X(n)^2-m*.15*g*(X(n)-X(n+1))=1/2*k*X(n+1)

but i don't know how to solve it. Many thanks in advance.
 
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  • #2


ok i redid my equation to 1/2*k*((x(n))^2-mg(-1^n(x(n))-(-1^(n+1))((x(n+1))=1/2*k*x(n+1)^2
 
  • #3


Based on your attempt at the solution, it seems like you have a good understanding of the concepts involved in this problem. To derive an expression for the maximum displacement of the mass at the end of each half oscillation as a function of the number of half oscillations, n, you can use the equation for the displacement of a mass-spring system:

x(t) = A*cos(ωt + φ)

Where:
A = amplitude of oscillation
ω = angular frequency (ω = √(k/m))
φ = phase angle

For a mass-spring system, the maximum displacement occurs at the amplitude, so we can rewrite the equation as:

x(t) = A*cos(ωt)

We can then use the equation for the period of a mass-spring system:

T = 2π√(m/k)

To find the time at which the maximum displacement occurs, we can set ωt = π/2 (since cos(π/2) = 0), and solve for t:

ωt = π/2
√(k/m)*t = π/2
t = π/2*√(m/k)

Now, we can substitute this value of t into our original equation for displacement at time t:

x(π/2*√(m/k)) = A*cos(π/2)
x(π/2*√(m/k)) = A*0
x(π/2*√(m/k)) = 0

Therefore, the maximum displacement, A, occurs at t = π/2*√(m/k). We can also express this in terms of the period, T, using the equation above:

t = π/2*√(m/k)
t = π/2*√(m/k)*√(k/m)*√(k/m)
t = π/2*T

So, the maximum displacement occurs at t = π/2*T, which is the halfway point of one full oscillation. Therefore, for every half oscillation, the maximum displacement is equal to the amplitude, A. And the amplitude, A, is given by:

A = x(0) = x(π/2*T) = .8m

So, the maximum displacement at the end of each half oscillation is equal to .8m, regardless of the number of half oscillations, n
 

1. What is a mass-spring system?

A mass-spring system is a physical system consisting of a mass attached to a spring, which is fixed at one end and free to move at the other. The mass-spring system is commonly used in physics experiments and real-life applications to study oscillatory motion and energy conservation.

2. How is the maximum displacement of a mass-spring system calculated?

The maximum displacement of a mass-spring system can be calculated using the formula x = A * sin(ωt), where A is the amplitude (maximum displacement from equilibrium) and ω is the angular frequency. Alternatively, it can also be calculated by finding the point where the kinetic energy of the mass is equal to the potential energy stored in the spring.

3. What factors affect the maximum displacement of a mass-spring system?

The maximum displacement of a mass-spring system is affected by several factors, including the mass of the object, the stiffness of the spring, and the amplitude of the oscillations. Additionally, external forces such as friction and air resistance can also affect the maximum displacement of the system.

4. How does changing the mass affect the maximum displacement of a mass-spring system?

Increasing the mass in a mass-spring system will result in a decrease in the maximum displacement, while decreasing the mass will result in an increase in the maximum displacement. This is because the mass affects the inertia of the system, which determines how much energy is required to move the mass and spring.

5. Can the maximum displacement of a mass-spring system be greater than the amplitude?

No, the maximum displacement of a mass-spring system cannot be greater than the amplitude. The amplitude represents the maximum value of the displacement from the equilibrium position, so the maximum displacement of the system cannot exceed this value. However, it is possible for the maximum displacement to be equal to the amplitude.

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