# Finding amplitude of oscillating spring; x(t) and v(t) given

1. Sep 24, 2016

### tensor0910

1. The problem statement, all variables and given/known data

At t = 0 seconds, the position of the oscillator is found to be -27.33 cm and the velocity of the oscillator is +0.744 m/s. What is the amplitude of the oscillation?

2. Relevant equations

position/time equation for oscillation: x(t) = Acos(wt + phi )
for velocity: v(t) = -Vmax sin(wt + phi )

3. The attempt at a solution

at t = 0, the position is -.2733 m and the velocity is .744 m/s.
This leaves us with two equations with 2 unknowns. ( I think ) this is how they should look:

-.2733 = Acos(phi)
.744 = -.744sin (phi )

I unsuccessfully attempted to solve for either unknown and ended up with:

tan(phi ) = -A/-.2733

This is where I got stuck. Not even sure if my algebra is correct. Any/all tips are appreciated. Thank you in advance!

Edit: Original problem coupled with question. Sorry, forum noob here :-)

A mass of 1.4 kg is attached to a single spring and set into oscillation. The period of the spring oscillation is measured to be 4.22 seconds

Last edited: Sep 24, 2016
2. Sep 24, 2016

### Buzz Bloom

Hi tensor:

I see two things that appear to be not quite right in your work.
1. You omitted the equation v(t) = dx/dt.
2. How did you arrive at the value of the coefficient of sin in the quoted equation?

Hope this helps.

Regards,
Buzz

3. Sep 24, 2016

### LCKurtz

If $x = \cos(\omega t + \phi)$ then $v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$, so you have three unknowns $A,~\omega,\text{ and }\phi$. Putting $t=0$ in those two gives two equations. And if you know the period is $4.22$ seconds, that should give you $\omega$. It might be helpful to use the $\sin^2 + \cos^2$ identity in your work.

4. Sep 24, 2016

### tensor0910

thank you for the reply Buzz!

After relooking at the problem with Buzz's tip I came up with these equations:

-.2733 = Acos( Φ ) x(t)
.744 = -ωAsin( Φ ) v(t)

I solved for ω using 2π/T ( calculator in rad ) and came up with 1.488

solved for A: A = -.2733/cosΦ

plugged A into v(t) and came up with :

.744 = (.40667)(tanΦ)

isolated tan(Φ) and solved for Φ and came up with...1.070.

But the program says its wrong!

Does anyone have any ideas?

Edit: solved for Amplitude using -.2733/cos (1.070.)

Last edited: Sep 24, 2016
5. Sep 24, 2016

### tensor0910

Figured it out. Calculator needs to be back in Deg. to finish problem. Φ = 61.338. Plug into x(t)

Thank you Buzz and LCKurtz for the help!

6. Sep 25, 2016

### ehild

The amplitude is asked, you do not need the angle.
-0.2733 = Acos( Φ ) = x(t)
0.744 = -ωAsin( Φ ) = v(t)
ω=1.489
You can write the second equation in form Asin( Φ ) = 0.744/ω. Square both this equation and the first one, add up the squares, and use @LCKurtz hint about sin2 + cos2.