Q2. A water wove, troveling in a straight line on a lake (the x-axis) has a tran

Question

Q2. A water wove, troveling in a straight line on a lake (the x-axis) has a transverse acceleration described (18 pts. toteal by ayfx, t): 3.50m/s cos2r[0.750 (8.00mx (0.833s)t] Answer the following questions regarding this water wave a. How much time, T, does it takes for one complete wave pattern to go past a fisherman in a boat at anchor? (2.0 pts.) b. What horizontal distance, d, does the crest of the wave travel in that time? (2.0 pts.) c. What is the amplitude, A, of the oscillation? Be careful here! (2.0 pts.) d, what is the phase angle.» nd what is the number of waves per second that pass the fisherman? (20 pts) e. How fast, vo, does a wave crest travel past the fisherman? (2.0 pts) f. What is the maximum transverse speed, vy, of his boat? (2.0 pts.) g, what is the acceleration, adx, t) of the wave at x : 0.100m and at 2.00s? (2.0 pts.) 9. formula for the wove'svertical displacement, ylx, t). explicitly in terms of k, , the values and units found above. Include the volues of all variables, except x and t. h Write a general (2.0 pts) ylx, t) i. If the omplitude, A, of the oscillation of the wove triplss, wheat is the change in frequency of the boat? (2.0 pts.)

Explanation / Answer

given equation

a = 3.5 cos(2*pi(0.75 + 8x + 0.833t))

comparing with

a = Aw^2 cos(kx + wt)

we get

k = 2*pi*8

w = 2*pi*0.833

a. hence time period T = 2*pi/w = 2*pi/2*pi*0.833 = 1.2004 s

b. horizontal distance = lambda = wavelength

lambda = 2*pi/k = 1/8 = 0.125 m

c. amplitude of osscilation = A

Aw^2 = 3.5

A = 0.1277 m

d. phase angle = 2*pi*0.75 rad = 4.71238 rad = 270 deg

numebr of waves per second = 1/T = 0.833 waves per second

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