Question

A massless spring with k = 172 N/m is hung from a support and a 1.19 kg mass is attached to the bottom of the spring. The position of the mass when it is in equilibrium is y = 0. If the mass is displaced below this point, we take its displacement to be positive. At t = 0, the mass has the initial conditions y(0) = 0.2 m and [dy/dt]t=0 = −1.1 m/s.

Its position at later times is
y(t) = A cos(ωt + δ) .

What is the value of A?

What is the value of ω?

What is the value of δ?

Answer 1

Holler if you need clarification on anything ^_^

Answer 2

So we have Newton and conservation of energy

\[
\\ \
\\ -ky = ma \hspace{85px} \textrm{Newton's law}
\\ \
\\ \ \ \ \ \ ma + ky = 0
\\ \
\\ \ \ \ \ \ m \left( \frac{\mathrm{d}^2 y}{\mathrm{d}t^2} \right) + ky = 0
\]
\[
\frac{1}{2}mv^2 + \frac{1}{2}ky^2 = E \hspace{35px} \textrm{Conservation of total mechanical energy}
\\ \
\\ \ \frac{1}{2}m \left( \frac{\mathrm{d} y}{\mathrm{d}t} \right)^2 + \frac{1}{2}ky^2 = E

\]
We know that our position is of the form

\[ y_{(t)} = A\cos(\omega t + \delta)
\\ \
\\ \textrm{So velocity is}
\\ \
\frac{\mathrm{d}y_{(t)}}{\mathrm{d}t} = -A \omega \sin(\omega t + \delta)
\\ \
\\ \ \textrm{and acceleration is}
\\ \
\\ \frac{\mathrm{d^2}y_{(t)}}{\mathrm{d}t^2} = -A \omega^2 \cos(\omega t + \delta)
\]


So from Newton's Laws we can see

\[ ma = -ky
\\ \
\\ \
m \left( \frac{\mathrm{d^2}y_{(t)}}{\mathrm{d}t^2} \right) = -ky
\\ \
\\ \
m \left( -A \omega^2 \cos(\omega t + \delta) \right) = -k(A\cos(\omega t + \delta))
\\ \
\\ \
\omega^2 = \frac{k}{m} \ \ \textrm{ angular velocity=omega }
\]


And similarly from Conservation of mechanical energy we glean
\[ \frac{1}{2}m \left( \frac{\mathrm{d} y}{\mathrm{d}t} \right)^2 + \frac{1}{2}ky^2 = E
\\ \
\\ \
\frac{1}{2}m \left(-A \omega \sin(\omega t + \delta)\right)^2 + \frac{1}{2}k( A\cos(\omega t + \delta))^2 = E
\\ \
\\ \
A = \sqrt{ \frac{2E}{k}}\]

Plugging in initial conditions gives you
\[ \frac{1}{2}m \left( \frac{\mathrm{d} y_{(0)}}{\mathrm{d}t} \right)^2 + \frac{1}{2}ky_{(0)}^2 = E_0
\\ \
\]
This gives you A
\[ A = \sqrt{ \frac{2E_0}{k}}
\]
Then you can solve for the phase, delta, using the initial position
\[ y_{(0)} = A\cos(\delta)\]

Answer 3

Thanks again you're like my physics savant lol . It's always starting these questions that I have a problem.

Answer 4

Welcome, and I know what you mean. There's a weird loopiness with that diff eqs that seems counter intuitive or something. For me at least ^_^

Answer 5

Also, omega is angular frequency, not angular velocity. :P

Answer 6

Don't worry I understood what you meant