a) To find the vertical displacement when the velocity is 0.8m/s, we need to find the first derivative of the given equation:
h(t) = 0.6cos(2t) + 0.8sin(t)
h'(t) = -1.2sin(2t) + 0.8cos(t)
Now, set h'(t) = 0.8:
-1.2sin(2t) + 0.8cos(t) = 0.8
-3(sin(2t) - cos(t)) = 2
sin(2t) - cos(t) = 2/3
Now, use the double angle identity, sin(2t) = 2sin(t)cos(t):
2sin(t)cos(t) - cos(t) = 2/3
cos(t)(2sin(t) - 1) = 2/3
For the wave's vertical displacement when the velocity is 0.8m/s, plug in the values of t that satisfy the above equation into the original equation h(t).
b) To find the maximum velocity, we need the second derivative of h(t):
h''(t) = -2.4cos(2t) - 0.8sin(t)
To find the critical points, set h''(t) = 0:
-2.4cos(2t) - 0.8sin(t) = 0
Now, find the values of t that make h''(t) = 0, and then plug them into the first derivative equation, h'(t), to find the maximum velocity. I made an error in part a, Vmax occurs when h''(t) = 0, not when h'(t) = 0.
c) The wave changes from a hill to a trough when the vertical displacement is 0, so we need to find when h(t) = 0:
0.6cos(2t) + 0.8sin(t) = 0
Divide by 0.6:
cos(2t) + (4/3)sin(t) = 0
Now, find the values of t that make h(t) = 0. The first value of t for which this happens is when the wave switches from a hill to a trough.
An oceanographer measured a set of sea waves during a storm and modelled the vertical displacement of waves in meters using the equation h(t)=0.6cos2t+0.8sint, where t is the time in seconds.
a) Determine the vertical displacement of the wave when the velocity is 0.8m/s
Ans: -1.2sin2t+0.8cost = 0.8
-2.4(sint)(cost)+0.8cost = 0.8
cost(-2.4sint+0.8) = 0.8
cost = 0.8
t = cos-1(0.8) OR -2.4sint +0.8 = 0.8
=0.6 t = 0
b) Determine the maximum velocity of the wave and when it occurs.
Ans: cost(-2.4sint+0.8)=0
therefore t= 1.5 and 0.3 and Vmax occurs at t=1.5s
c) When does the wave first change from a hill to a trough? Explain.
Please check the above answers and help if they are incorrect, and need guidance with part c, is it asking for the height?
1 answer
1 answer