Amplitude of oscillating spring

[hopperd]

Homework Statement

a 185g mass attached to a horizontal spring oscilates at 5.2Hz. At t=0s, the amss is at x=5.80 and has a vx=-40.0cm. determine the amplitude

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

welcome to pf!

hi hopperd! welcome to pf!

it's shm, so write a differential equation, and solve it with the given initial conditions …

what do you get? ​

 

[hopperd]

Thanks,
Thats the problem. I have tried 3 diff equations, and get the same answer, but it's the wrong one. So I must be missing something. What equation would you use?

 

[tiny-tim]

hi hopperd!

show us your full calculations, and the official answer, and then we'll see what went wrong, and we'll know how to help!

 

[paranormal]

The amplitude of an oscillating spring is the maximum displacement of the mass from its equilibrium point. In this case, the amplitude can be calculated by using the given information of the mass, frequency, and initial position and velocity of the mass.

Using the equation for the frequency of a spring-mass system, f = 1/2π√(k/m), where k is the spring constant and m is the mass, we can solve for k. Plugging in the given values, we get k = (2πf)^2m = (2π*5.2Hz)^2*0.185kg = 6.071 N/m.

Next, we can use the equation for the displacement of a spring-mass system, x = A*cos(2πft), where A is the amplitude and t is time. At t = 0s, x = 5.80cm, so we can solve for A: 5.80cm = A*cos(0), therefore A = 5.80cm.

Finally, we can use the equation for the velocity of a spring-mass system, v = -2πfA*sin(2πft), to solve for the amplitude again. At t = 0s, v = -40.0cm/s, so we get -40.0cm/s = -2π*5.2Hz*A*sin(0), therefore A = 3.85cm.

Thus, the amplitude of the oscillating spring is 3.85cm.