Example (Syllabus) The deck of a ship was 2.4 m below the level of the wharf at low tide and 0.6 m above the level at high tide. Low tide was at 8:30 am and high tide was at 2.35 pm. Find when the deck was level with the wharf, if the motion was simple harmonic. Solution The most important part of answering this question is constructing a good scientific diagram of the physical situation. Take the initial conditions at the 8:30 am low tide = 0 s = 0 m.s-1 = -xmax = -1.5 m t v x The displacement as a function of time is At time = 0 t Hence The time interval from low tide to high tide is half-period T/2 8:30 am to 2:35 pm = 6 5 minutes = (6)(60)(60)+(5)(60) s = 21900 s t period = 43800 s T angular velocity 1.4345x10-4 rad.s-1 = We want to find the time when the deck is level with the wharf = ? s = 0.9 m position of wharf above equilibrium position (x = 0) t x = 1.5426x104 s = 4.2877 h = 4 h 17 m t The deck will be level with the wharf at time 12:47 pm
Example (Syllabus) The deck of a ship was 2.4 m below the level of the wharf at low tide and 0.6 m above the level at high tide. Low tide was at 8:30 am and high tide was at 2.35 pm. Find when the deck was level with the wharf, if the motion was simple harmonic. Solution The most important part of answering this question is constructing a good scientific diagram of the physical situation. Take the initial conditions at the 8:30 am low tide t = 0 s v = 0 m.s-1 x = -xmax = -1.5 m The displacement as a function of time is At time t = 0 Hence The time interval from low tide to high tide is half-period T/2 8:30 am to 2:35 pm t = 6 5 minutes = (6)(60)(60)+(5)(60) s = 21900 s period T = 43800 s angular velocity = 1.4345x10-4 rad.s-1 We want to find the time when the deck is level with the wharf t = ? s x = 0.9 m position of wharf above equilibrium position (x = 0) t = 1.5426x104 s = 4.2877 h = 4 h 17 m The deck will be level with the wharf at time 12:47 pm