Find: L [ft] L L d A) 25 B) 144 C) 322 D) 864 T = 13 [s] d = 20 [ft] Find the wavelength, in feet. [pause] In this problem, --- T = 13 [s] d = 20 [ft]

Find: L [ft] L L d A) 25 B) 144 C) 322 D) 864 T = 13 [s] d = 20 [ft] we are provided the period of the wave, T, and the depth of water, ---- T = 13 [s] d = 20 [ft]

Find: L [ft] L L d A) 25 B) 144 C) 322 D) 864 T = 13 [s] d = 20 [ft] d. [pause] To find the wavelength, L, we’ll divide d, --- T = 13 [s] d = 20 [ft]

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = d/L by the quotient d over L. From the problem statement, we already know the depth, equals, --- L = d/L

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = d/L 20 feet. [pause]. The quotient d over L can be determined based on the quotient, --- L = d/L

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = d/L based on d/Lo d over, L knot. L knot refers to the wavelength --- L = d/L based on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = d/L based deep water for deep water conditions, which occurs when, d over L, --- L = d/L based deep water on d/Lo conditions

Find: L [ft] ≥ L L d d = 20 [ft] T = 13 [s] d L = d/L based deep water is at least, 0.50. Since our values of d and T may not produce --- L = d/L based deep water d ≥ 0.50 on d/Lo conditions L

? Find: L [ft] ≥ L L d d = 20 [ft] T = 13 [s] d L = d/L based deep water conditions, we’ll need to look up the quotient, --- L = ? d/L based deep water d ≥ 0.50 on d/Lo conditions L

? Find: L [ft] ≥ L L d d = 20 [ft] T = 13 [s] d L = d/L based d over L. [pause] L knot equals, --- L = ? d/L based deep water d ≥ 0.50 on d/Lo conditions L

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = d/L g * T2 based Lo = g T squared, divided by 2 PI. [pause] This equation for L knot is a simplified version of our equation for L, --- L = d/L g * T2 based Lo = 2 * π on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] g * T2 d L = L = because for deep water conditions, the term hyperbolic tangent of k d --- L = L = tanh ( k * d ) 2 * π * d/L g * T2 based Lo = 2 * π on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] g * T2 d L = L = equals 1, and can be removed from the equation. [pause] In this equation, g, --- L = L = tanh ( k * d ) 2 * π * d/L g * T2 tanh ( k * d )=1 based Lo = 2 * π on d/Lo when computing Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] g * T2 d L = L = is the gravitational acceleration constant, which equals, --- L = L = tanh ( k * d ) 2 * π * d/L g * T2 gravitational based Lo = 2 * π acceleration on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = g=32.16 [ft/s2] d/L 32.16 feet per second squared. [pause] Variable T represents the wave period, --- L = g=32.16 [ft/s2] d/L g * T2 gravitational based Lo = 2 * π acceleration on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d L = g=32.16 [ft/s2] d/L which equals 13 seconds. [pause] This makes L knot equal to, --- L = g=32.16 [ft/s2] d/L g * T2 based Lo = 2 * π wave period on d/Lo

Find: L [ft] = 865.01 [ft] L L d d = 20 [ft] T = 13 [s] d L = 865.01 feet. [pause] Now we can compute --- L = g=32.16 [ft/s2] d/L g * T2 based Lo = = 865.01 [ft] 2 * π on d/Lo

Find: L [ft] = 865.01 [ft] L L d d = 20 [ft] T = 13 [s] d d L = d over L knot, which is 20 feet --- L = g=32.16 [ft/s2] d/L Lo g * T2 based Lo = = 865.01 [ft] 2 * π on d/Lo

Find: L [ft] = 865.01 [ft] L L d d = 20 [ft] T = 13 [s] d d L = divided by, 865.01 feet, which equals, --- L = g=32.16 [ft/s2] d/L Lo g * T2 based Lo = = 865.01 [ft] 2 * π on d/Lo

Find: L [ft] = 0.02312 = 865.01 [ft] L L d d = 20 [ft] T = 13 [s] d d 0.02312. [pause] In the look up tables, we can determine the quotient d over L --- L = g=32.16 [ft/s2] = 0.02312 d/L Lo g * T2 based Lo = = 865.01 [ft] 2 * π on d/Lo

Find: L [ft] = 0.02312 = 865.01 [ft] L L d d = 20 [ft] T = 13 [s] d d by interpolating between, d over L knot values of, ---- L = g=32.16 [ft/s2] = 0.02312 d/L Lo g * T2 based Lo = = 865.01 [ft] 2 * π on d/Lo

Find: L [ft] L L d d = 20 [ft] T = 13 [s] d/Lo d/L d L = 0.023 0.06200 0.023 and 0.024. After writing out the equation to solve for d over L, --- L = 0.023 0.06200 d/L 0.024 0.06340 d/Lo= 0.02312 d/L= ?

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d = 20 [ft] T = 13 [s] d/Lo d/L d using linear interpolation, we compute d over L, equals, --- L = 0.023 0.06200 d/L 0.024 0.06340 d/Lo= 0.02312 d/L= ?

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d/L = 0.06217 d = 20 [ft] d 0.06217. [pause] Lastly, we’ll plug this quotient into our equation for L, --- L = d/L d/Lo d/L d/Lo= 0.02312 0.023 0.06200 0.024 0.06340

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d/L = 0.06217 d = 20 [ft] d and the wavelength equals 20 feet divided by 0.06217, which equals, --- L = d/L d/Lo d/L d/Lo= 0.02312 0.023 0.06200 0.024 0.06340

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d/L = 0.06217 L = 321.70 [ft] d = 20 [ft] d 321.70 feet. [pause] L = d/L d/Lo d/L d/Lo= 0.02312 0.023 0.06200 0.024 0.06340

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d/L = 0.06217 L = 321.70 [ft] d = 20 [ft] A) 25 B) 144 C) 322 D) 864 d When looking over the possible solutions, --- L = d/L d/Lo d/Lo= 0.02312 0.023 0.024

Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023) ( 0.06340 - 0.06200 ) * d/L = 0.06217 L = 321.70 [ft] d = 20 [ft] A) 25 B) 144 C) 322 D) 864 d the correct answer is C. [fin] L = d/L answerC d/Lo= 0.02312