Find max energy stored in oscillating system

[smillphysics]

A 0.90 kg mass on a spring oscillates horizontally with little friction according to the following equation: x = 0.090cos(2.90t), where x is in meters and t in seconds. Find the maximum energy stored in the spring during an oscillation.

X=a*cos (omega*t)
omega =2pi*frequency = sqrot k/m
x=mg/k which came from kx=mg

I got k=7.569 from omega=sqrt m/k, then I plugged k into kx=mg and got x=1.165
from here I would like to use well...im unsure. is the total energy equal to 1/2kx^2 because I have x and k if they are correct.

 

[smillphysics]

Part B of the problem is to
Find the maximum velocity of the mass. Would I use v=-Aomega sin (omega *t)?

 

[nlightner]

First, we need to find the frequency of oscillation using the given equation: x = 0.090cos(2.90t). The coefficient of t, 2.90, is the frequency in Hz. This means that the system oscillates 2.90 times per second.

Next, we can use the equation for potential energy in a spring system: U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium.

Using the given values, we can calculate the spring constant: k = (2pi*f)^2*m = (2pi*2.90)^2*0.90 = 66.96 N/m.

To find the maximum energy stored in the spring, we need to know the maximum displacement from equilibrium, which is the amplitude of the oscillation. From the given equation, we can see that the amplitude is 0.090 m.

Now, we can plug in our values into the potential energy equation: U = 1/2*66.96*(0.090)^2 = 0.027 J.

Therefore, the maximum energy stored in the spring during an oscillation is 0.027 J.