Finding amplitude of oscillator

[natasgan]

Homework Statement

a 200g mass attached to a horizontal spring oscillates at a frequency of 2.0 Hz. At t=0s, the mass is at x=5.0 cm and has v=-30 cm/s. Determine:
a) period
b)angular frequency
c) AMPLITUDE

Homework Equations

x(t)=Acos(wt+psi)
v(t)=-Awsin(wt+psi)
w=2*pi*f
T=1/f

The Attempt at a Solution

I have solved for the period=1/2 s
I have also solved for angular frequency=2*pi*2=4*pi
However, I am confused about solving for amplitude. I thought that I could plug in 0.05 m and 0s to the x(t) formula, but I do not know psi (phase constant) so I am stuck.

 

[Ofey]

When solving for the amplitude, try using conservation of energy.

--> All energy in the system is potential energy when the spring is extended the most. (When the spring is extended the most, the length extended is the amplitude)

The spring constant you can get from

T=2*pi*sqrt(m/k)

 

[nlightner]

The amplitude of an oscillator is the maximum displacement from the equilibrium position. In this case, the equilibrium position is at x=0, so the amplitude can be found by measuring the maximum displacement from x=0.

To find the amplitude, we can use the given information about the initial position and velocity of the mass.

At t=0s, the mass is at x=5.0 cm. This means that the amplitude is 5.0 cm, or 0.05 m.

Another way to find the amplitude is by using the equation for velocity: v(t)=-Awsin(wt+psi). At t=0s, v=-30 cm/s, so we can plug in these values and solve for A:

-30 cm/s=-A*4*pi*sin(0+psi)
-30 cm/s=-4*pi*A*sin(psi)
A=0.05 m

Therefore, the amplitude of the oscillator is 0.05 m.