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?
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.
Your solutions for parts a) and b) are correct.
For part c), the question is asking for the time at which the wave changes from a hill (crest) to a trough (valley). To determine when this change occurs, we need to find the points where the wave crosses the x-axis (zero displacement).
To find these points, we set the equation h(t) equal to zero and solve for t:
0.6cos2t + 0.8sin(t) = 0
We can simplify this equation by dividing through by 0.8:
0.75cos2t + sin(t) = 0
Now, we can use a graphing calculator or tool to find the values of t where this equation is satisfied. The solutions will correspond to the times when the wave changes from a hill to a trough.
Alternatively, we can use trigonometric identities to simplify the equation further:
0.75(1 - 2sin^2(t)) + sin(t) = 0
0.75 - 1.5sin^2(t) + sin(t) = 0
Rearranging the terms:
1.5sin^2(t) - sin(t) - 0.75 = 0
Now, we can solve this quadratic equation for sin(t) using factoring or the quadratic formula:
sin(t) = (1 ± √(1 + 4(1.5)(0.75))) / (2(1.5))
sin(t) = (1 ± √(1 + 4(1.5)(0.75))) / 3
Compute the values of sin(t), and then use the inverse sine function to find the values of t. The solutions will correspond to the times when the wave changes from a hill to a trough.
Please note that since this is a complicated trigonometric equation and involves special angles, it is recommended to use a graphing calculator or tool for accurate and efficient calculation of the solutions.
Let's go through each part and check the answers:
a) To determine the vertical displacement of the wave when the velocity is 0.8 m/s, we need to solve the equation h(t) = 0.6cos(2t) + 0.8sin(t) = 0.8. Your working is correct up until this point.
We have -1.2sin(2t) + 0.8cos(t) = 0.8. Now, if we rearrange the equation to isolate the sine term, we get: -1.2sin(2t) = 0.8 - 0.8cos(t).
Simplifying further, we have -1.2sin(2t) = 0.8(1 - cos(t)).
Now, since cos(t) varies between -1 and 1, we know that 1 - cos(t) varies between 0 and 2. Therefore, for the right-hand side to equal 0.8, we must have sin(2t) = -0.8/1.2 = -2/3.
As sin(theta) varies between -1 and 1, we cannot achieve sin(2t) = -2/3 with real values of theta. So, there is no solution for t in this case, and therefore, there is no time at which the vertical displacement is 0.8 m.
b) To determine the maximum velocity of the wave and when it occurs, we need to find the first derivative of the equation h(t) = 0.6cos(2t) + 0.8sin(t). Taking the derivative with respect to t, we get dh(t)/dt = -1.2sin(2t) + 0.8cos(t).
To find the maximum velocity, we need to find the time(s) at which dh(t)/dt = 0. Setting the derivative equal to 0, we have -1.2sin(2t) + 0.8cos(t) = 0.
Simplifying this equation, we can write it as cos(t) = 1.2sin(2t). By using the identity sin(2t) = 2sin(t)cos(t), we can rewrite it as cos(t) = 2.4sin(t)cos(t).
Now, dividing both sides of the equation by cos(t) (since cos(t) cannot be zero), we have 1 = 2.4sin(t). Hence, sin(t) = 1/2.4 = 5/12.
Now, we know that sin(theta) = 1/2 when theta = pi/6 or 5pi/6. Therefore, the times at which the maximum velocity occurs are t = pi/6 and t = 5pi/6.
c) To determine when the wave first changes from a hill to a trough, we need to find the times at which the vertical displacement goes from positive to negative.
The equation h(t) = 0.6cos(2t) + 0.8sin(t) represents the vertical displacement. To find the points where this changes sign, we need to find when h(t) = 0.
So, we have 0.6cos(2t) + 0.8sin(t) = 0. To solve this equation, we need to use numerical methods or graphing techniques. Based on the graph of h(t), we can observe the points at which the wave changes from a hill to a trough.
It is not explicitly asking for the height, but to find when the wave changes from a hill to a trough, we need to find the times at which the vertical displacement is zero.
I hope this explanation helps clarify the answers and provides guidance for part c!