I came across a physics problem for homework that I was able to work out on my own, however when I got to class, my teacher said that my numbers were not what the book had calculated, but couldn't tell if there was a flaw in my logic that would prevent me from getting the correct phi and therefore the correct answers.(adsbygoogle = window.adsbygoogle || []).push({});

The question:

A simple harmonic oscillator consists of a block of mass 4.20 kg attached to a spring of spring constant 200 N/m. When t = 4.00 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s.

My approach

a.) What is the amplitude of the oscillations?

I used conservation of mechanical energy to solve this part of the problem, and got the correct amplitude this way:

(1/2)kx^2 + (1/2)mv^2 = (1/2)k(amplitude)^2

amplitude = .5114 m

b.) What was the position of the mass at t = 0 s?

For this part, I plugged the position I was given and the time I was given as well as the amplitude and omega, which I solved for, into this formula:

x(t) = amplitude * cos(omega * t + phi)

I solved for omega:

omega = (k/m)^(1/2)

omega = 6.9 rad/s

and used the amplitude from part a) and the given x and t to solve for phi

.129 = .5114 * cos(6.9 * 4 + phi)

.252 = cos(27.6 + phi)

1.316 = 27.6 + phi

phi = -26.28 rad

then I use this phi in the x(t) function to solve for x at t = 0:

x(0) = .5114 * cos(-26.28)

x(0) = .21 m

c.) What was the velocity of the mass at t = 0 s?

I use the same phi, amp, and omega from before, except in the equation:

v(t) = -omega * amplitude * sin(omega * t + phi)

v(0) = -(6.9) * (.5114) * sin(-26.28)

v(0) = 3.217 m/s

When my teacher does the problem another way, however, he got a phi that was different from mine.

Teacher's Approach

v(4)/x(4) = [-omega * amplitude * sin(omega * t + phi)] / [amplitude * cos(omega * t + phi)]

v(4)/x(4) = -omega * tan(omega * t + phi)

26.473 = -6.9 * tan(6.9 * 4 + phi)

-3.837 = tan(27.6 + phi)

-1.316 = 27.6 + phi

phi = -28.916 rad

He then proceeded to solve for the amplitude using x(t) = amplitude * cos(omega * t + phi) and x(0) and v(0) in pretty much the same way that I did.

My question is, why did using his method get the equation:

-1.316 = 27.6 + phi

phi = -28.916 rad

when my way instead gave me:

1.316 = 27.6 + phi

phi = -26.28 rad

?

In order to get the phi that my teacher got, I would have to alter the formula that I used to solve for phi to:

x(t) = amplitude * cos(-omega * t - phi)

which I don't think is correct. I would greatly appreciate any insight into this matter.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Simple Harmonic Motion homework

**Physics Forums | Science Articles, Homework Help, Discussion**