# At a certain harbor, the tides cause the ocean surface to rise and fall in simple harmonic motion, with a period of 12.5 h. How long does it take for the water to fall from its maximum height to one-half its maximum height above its average (equilibrium) level?

Jitender Pal
8 years ago
To obtain t in terms of w, substitute w = 2π/T in the equation t = π/3w,
t = π/3w
= π/3(2π/T)
= T/6
To find out t (time), substitute 12.5 h for period T in the equation t = T/6,
t = T/6
=12.5 h/6
=2.08 h
From the above observation we conclude that, it will take 2.08 h for the water to fall from its maximum height to one-half its maximum height above its average (equilibrium) level.
Jitender Pal
8 years ago
To obtain t in terms of w, substitute w = 2π/T in the equation t = π/3w,
t = π/3w
= π/3(2π/T)
= T/6
To find out t (time), substitute 12.5 h for period T in the equation t = T/6,
t = T/6
=12.5 h/6
=2.08 h
From the above observation we conclude that, it will take 2.08 h for the water to fall from its maximum height to one-half its maximum height above its average (equilibrium) level.

pa1
357 Points
8 years ago
 To obtain t in terms of w, substitute w = 2π/T in the equation t = π/3w, t = π/3w = π/3(2π/T) = T/6 To find out t (time), substitute 12.5 h for period T in the equation t =...
pa1
357 Points
8 years ago
To obtain t in terms of w, substitute w = 2π/T in the equation t = π/3w, t = π/3w = π/3(2π/T) = T/6 To find out t (time), substitute 12.5 h for period T in the equation t =...