Answer
(a) The wave is moving in the negative x-direction.
(b) $A = 2.0~mm$
$\omega = 1570~rad/s$
$k = 157~rad/m$
(c) The three smallest non-negative times that this snapshot could have been taken are $1.0~ms$, $5.0~ms$, and $9.0~ms$
Work Step by Step
(a) From the term $(\omega~t+k~x)$, we can see that the wave is moving in the negative x-direction.
(b) The points oscillate between $y = -2.0~mm$ and $y = 2.0~mm$, so the amplitude is $A = 2.0~mm$
On the graph, we can see that one cycle has a length of $4.0~cm$, so $\lambda = 4.0~cm$.
We can find $\omega$:
$\omega = 2\pi~f = 2\pi~\frac{v}{\lambda} = (2\pi)~\frac{10.0~m/s}{0.040~m} = 1570~rad/s$
We can find $k$:
$k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.040~m} = 157~rad/m$
(c) $y(0,0) = A~cos(\omega~(0)+k~(0)) = A$
At $t=0$, the point $x = 0$ would be located at a maximum amplitude of $y = 2.0~mm$. Since the wave is moving in the negative x-direction, the snapshot shows the wave for some time $nT+\frac{T}{4}$, for some integer $n$, since the initial position has been shifted to the left by $\frac{1}{4}$ of a cycle.
We can find the period $T$:
$T = \frac{1}{f} = \frac{\lambda}{v} = \frac{0.040~m}{10.0~m/s} = 0.0040~s = 4.0~ms$
We can find the three smallest non-negative times that this snapshot could have been taken:
$t_1 = (0)(T)+\frac{T}{4} = (0)(4.0~ms)+\frac{4.0~ms}{4} = 1.0~ms$
$t_2 = (1)(T)+\frac{T}{4} = (1)(4.0~ms)+\frac{4.0~ms}{4} = 5.0~ms$
$t_3 = (2)(T)+\frac{T}{4} = (2)(4.0~ms)+\frac{4.0~ms}{4} = 9.0~ms$
The three smallest non-negative times that this snapshot could have been taken are $1.0~ms$, $5.0~ms$, and $9.0~ms$
