1. 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]

2. 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]

3. 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]

4. 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

5. 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

6. 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

7. 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

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

9. ? 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

10. ? 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

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

18. 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 =
= [ft]
2 * π
on d/Lo

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

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

21. 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] =
d/L Lo
g * T2
based
Lo =
= [ft]
2 * π
on d/Lo

22. 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 After writing out the equation to solve for d over L, ---
L =
0.023
d/L
0.024
d/Lo=
d/L= ?

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

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

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

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

27. Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023)
( ) *
d/L =
L = [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=
0.023
0.024

28. Find: L [ft] + * (0.02312 - 0.023) d/L = 0.06200 (0.024 - 0.023)
( ) *
d/L =
L = [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=