Solve Mass Using Spring & Oscillation

[azn4life1990]

Homework Statement

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure

http://session.masteringphysics.com/problemAsset/1073872/3/knight_Figure_14_36.jpg
What is her mass if the spring constant is 240N/m

Homework Equations

Hookes Law
F = -kx
U = (1/2)*k*x²

The Attempt at a Solution

So the spring was pulled back 1.4 metres and then released where it went to the point 0.6
so 0.6 is the point at equilibrium, so -0.8 is the displacement and that is the x

F = (240 * 0.8)
and then I am not sure what to do, I am not even sure if I am doing this right and help would be appreciated

 

[azn4life1990]

im pretty sure my approach is wrong...

F = -260 * -0.8 = 208N

mg = weight

208/9.8 = m

m = 21.224489795918367346938775510204kg

im sure that's wrong is there a better way to approach this problem and why is the way I am approaching it wrong?

 

[nlightner]

I would approach this problem by first understanding the concept of oscillation and how it relates to mass and spring constants. Oscillation is the back and forth movement of an object around an equilibrium point, and in this case, the object is the astronaut attached to the spring.

Using Hookes Law, we know that the force exerted by the spring is directly proportional to the displacement of the object from its equilibrium position. This can be expressed as F = -kx, where k is the spring constant and x is the displacement. In this case, we are given the spring constant as 240N/m and the displacement of the astronaut as -0.8m.

To solve for the mass of the astronaut, we can use the equation for potential energy of a spring, U = (1/2)*k*x^2. We know that at the equilibrium point, the potential energy is at its minimum, which means that all of the energy is in the form of kinetic energy. Therefore, we can equate the kinetic energy of the astronaut to the potential energy of the spring at the equilibrium point.

So, we have (1/2)mv^2 = (1/2)kx^2, where m is the mass of the astronaut, v is the velocity, k is the spring constant, and x is the displacement. We can rearrange this equation to solve for the mass, m = (kx^2)/v^2.

Now, we need to find the velocity of the astronaut at the equilibrium point. We can use the equation for simple harmonic motion, T = 2π√(m/k), where T is the period of oscillation. We can rearrange this equation to solve for the velocity, v = 2πx/T.

Plugging this into our previous equation, we get m = (kx^2)/(4π^2x^2/T^2). Simplifying this, we get m = (kT^2)/(4π^2).

Using the given information, we can calculate the period of oscillation, T = 4.8s, and the spring constant, k = 240N/m. Plugging these values into our equation, we get m = (240N/m)(4.8s)^2/(4π^2) = 10.7kg.

Therefore, the mass of the astronaut is 10.7kg. This method