Solving a Horizontally Oscillating Ball Problem

[Cfem]

Homework Statement

A 130 g ball attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 17 cm/s when x = 4.2 cm

Find the:
-Amplitude of Oscillation
-Max acceleration
-Speed of the ball when x = 2.8cm

Homework Equations

T = 2*pi*(m/k)^1/2
Sinusoidal equations for position and velocity of a particle in SHM

The Attempt at a Solution

I'm not entirely sure where to start. I solved the velocity and position equations for t and set them equal to solve for A, but finding an intersection for that is beyond me. Not sure how else to work the equation.

I figured if I could find A then I could take the derivative of the velocity equation and try to solve for the maximum acceleration, but that still leaves the phase shift and doesn't give an acceleration value to set it equal to.

Not sure where to begin on this one at all.

 

[novop]

I think using energy equations is the easiest way to solve this. The total energy is given by $$1/2kA^2 = 1/2kx^2 + 1/2mv^2$$

Solve for amplitude A.

As for the max acceleration, this will occur intuitively when all energy is potential and being converted into kinetic. Use $$F=ma$$ where$$F=-kx$$.

Hope that helps.

 

[Cfem]

I hadn't thought about the energy perspective. Thanks for that one.

But about finding the max acceleration...

Since the potential of a spring is given by (1/2)(k)(x^2), I figure it was intuitive that the max acceleration would be when the spring is stretch to its max distance, its amplitude.

But (1/2)(3)(5.5^2) = (.13)(a) gives me a huge value for acceleration, which is being spit back. So I'm wondering what I'm missing. (yes, 5.5cm is the right amplitude).

 

[novop]

I don't think I made myself clear before. I meant to solve the equation $$kx = ma$$ (where x is the amplitude A) by equating forces. Your intuition is right, though.

 

[rdl]

As a scientist, the first step in solving this problem would be to identify the known and unknown variables. In this case, the known variables are the mass of the ball (130 g), the spring constant (3.0 N/m), the initial velocity (17 cm/s), and the position (x = 4.2 cm). The unknown variables are the amplitude of oscillation, the maximum acceleration, and the speed of the ball at x = 2.8 cm.

Next, we can use the equations for simple harmonic motion (SHM) to solve for the unknown variables. The equation for the period of oscillation (T) is T = 2*pi*(m/k)^1/2, where m is the mass of the ball and k is the spring constant. Substituting the known values into this equation, we get T = 2*pi*(0.130 kg/3.0 N/m)^1/2 = 0.965 seconds.

The amplitude of oscillation (A) can be found using the equation for position in SHM, x = A*sin(2*pi*t/T), where t is the time and T is the period. Since we know the position at t = 0 (x = 4.2 cm), we can solve for A by setting t = 0 and solving for A. This gives us A = 4.2 cm.

To find the maximum acceleration, we can use the equation for acceleration in SHM, a = -A*(2*pi/T)^2*cos(2*pi*t/T), where A is the amplitude and T is the period. Substituting the known values, we get a = -(4.2 cm)*(2*pi/0.965 s)^2*cos(2*pi*t/0.965 s). Since we are looking for the maximum acceleration, we can set t = 0 and solve for a, which gives us a = 0.173 m/s^2.

Finally, to find the speed of the ball at x = 2.8 cm, we can use the equation for velocity in SHM, v = 2*pi*A*cos(2*pi*t/T), where A is the amplitude and T is the period. Substituting the known values, we get v = 2*pi*(4.2 cm)*cos(2*pi*t/0.965 s). Setting t = 0 and solving for v, we get v