Question

A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat

Answer 1

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

$v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}$

the relative velocity is:

$v'=13.25m/s-5m/s=8.25\frac{m}{s}$

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

$x'=v't=(8.25m/s)(20s)=165m$

Next, you use the general for of a wave:

$f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)$

you take the amplitude as 2.0/2 = 1.0m.

$\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}$

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

$f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m$