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A certain bay with very high tides displays the following behavior. In one 12-h period the water starts at mean sea level, rises to 19 ft above, drops to 19 ft below, then returns to mean sea level. Assuming that the motion of the tides is simple harmonic, find an equation that describes the height of the tide in this bay above mean sea level. (Let y be the height above sea level in feet, and t the number of hours since the start of the 12-h period.) Sketch a graph that shows the level of the tides over a 12-h period y (feet) y (feet) 19 19 t (hours) t (hours) 12 3 12 19 19 y (feet) y (feet) 19 19 t (hours) t (hours) -19 -19 Need Help?Read tWatch Watch It

Answer 1

In one 12h period the water starts at mean sea level, rises 19 ft above, drops to 19 ft below, then returns to mean sea level. The motion of the tide is simple harmonic.

Since tide starts from the mean sea level, so it is sine wave. From the above data it is clear that amplitude of the oscillation of y, A=19 ft. ​​​​​​

In one 12h period the water starts at mean sea level, rises 19 ft above, drops to 19 ft below, then returns to mean sea level. The motion of the tide is simple harmonic. Since tide starts from the mean sea level, so it is sine wave. From the above data it is clear that amplitude of the oscillation of y, A = 19 ft. and time period of the oscillator T = 12 h Therefore T = 12 times 60 times 60 s = 43200 s Therefore Angular frequency w = 2 pi/T Therefore w = 2 times 3.14/43200 rad/s = 1.453703704 times 10^-4 rad/s We know that general expressions of a sine wave with no phase difference, is given by y = A sin wt In our case, equations of tide is given by y = 19 sin (1.454 times 10^-4 t) ft Second part: The graph- option 4th