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 21 ft above, drops to 21 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.)

the required equation is; y = 21 sin(πt/6)

Step-by-step explanation:

Given the data in the question;

Water rises above sea level = 21 ft

Water drops below sea level = 21 ft

so

maximum = 21 ft and also

minimum = 21 ft

Amplitude = ( maximum + minimum ) / 2

Amplitude = ( 21 + 21 ) / 2

Amplitude = ( 21 + 21 ) / 2 = 42/2 = 21 ft

Period = 12 hours

and we know that; period = 2π/b

so

12 = 2π/b

12b = 2π

b = 2π / 12

b = π/6

Standard equation for simple harmonic is; y = asin(bt)

we substitute

y = 21 sin(πt/6)

Hence the required equation is; y = 21 sin(πt/6)

16.9 times 2.85

step-by-step explanation: