Answer
(a) The amplitude is $4.0~mm$
(b) The wavelength is $1.05~m$
(c) The period is $0.0105~s$
(d) The wave speed is $100~m/s$
(e) The wave travels in the positive x-direction.
Work Step by Step
(a) In general: $y(x,t) = A~sin(\omega~t-k~x)$
We can see from the equation that $A = 4.0~mm$
The amplitude is $4.0~mm$
(b) We can find the wavelength:
$\lambda = \frac{2\pi}{k}$
$\lambda = \frac{2\pi}{6.0~rad/m}$
$\lambda = 1.05~m$
The wavelength is $1.05~m$
(c) We can find the period:
$T = \frac{2\pi}{\omega}$
$T = \frac{2\pi}{6.0\times 10^2~rad/s}$
$T = 0.0105~s$
The period is $0.0105~s$
(d) We can find the wave speed:
$v = \lambda~f$
$v = \lambda~\frac{\omega}{2\pi}$
$v = (1.05~m)~\frac{6.0\times 10^2~rad/s}{2\pi}$
$v = 100~m/s$
The wave speed is $100~m/s$
(e) From the term $(\omega~t-k~x)$, we can see that the wave travels in the positive x-direction.
