The problem given in the image asks for a mathematical model of a particle's vertical motion relative to the water's surface as it travels in and out of the water at a constant speed. Let's break it down:

Information from the problem:

• The particle reaches a maximum height of 1.3 m above the water level.
• The particle reaches a depth of 2.6 m below the water level.
• It starts at a depth of $\frac{13}{8} \, \text{m}$ below the water level.
• It takes $\frac{4}{3} \pi \, \text{seconds}$ to move from its starting depth, reach the maximum height, go down to the minimum depth, and return to the starting depth.

You need to write down a model for the height $h_m$ of the particle relative to the water level, in the form:

$h_m = A \sin(Bt + C) + D$

Where $A$, $B$, $C$, and $D$ are constants to be determined.

Step-by-step solution:

1. Amplitude ($A$):

   • The particle moves between a maximum height of 1.3 m and a minimum depth of 2.6 m below the water. This gives a total vertical range of $1.3 + 2.6 = 3.9 \, \text{m}$.
   • The amplitude $A$ is half of this total range, so: $A = \frac{3.9}{2} = 1.95$

2. Vertical shift ($D$):

   • Since the particle moves both above and below the water level, the vertical shift is the average of the maximum height and the minimum depth. $D = \frac{1.3 + (-2.6)}{2} = -0.65$

3. Period and B:

   • The period of the motion is the time it takes for the particle to complete one full cycle (from the starting depth to the maximum height, to the minimum depth, and back to the starting depth). The problem states this is $\frac{4}{3} \pi \, \text{seconds}$.
   • The formula for the period of a sine function is $\frac{2\pi}{B}$, so we can set this equal to $\frac{4}{3} \pi$: $\frac{2\pi}{B} = \frac{4}{3} \pi$ Solving for $B$: $B = \frac{3}{2}$

4. Phase shift ($C$):

   • The phase shift $C$ depends on the initial position of the particle. The particle starts at a depth of $\frac{13}{8} \, \text{m}$ below the water level, or $-\frac{13}{8} \, \text{m}$.
   • To determine $C$, we plug the initial time $t = 0$ into the equation and set the height equal to the starting depth $h_m = -\frac{13}{8}$.

The equation is: $h_m(0) = A \sin(C) + D = -\frac{13}{8}$ Substitute the values for $A$ and $D$: $1.95 \sin(C) - 0.65 = -\frac{13}{8}$ Convert $-\frac{13}{8}$ to a decimal: $1.95 \sin(C) - 0.65 = -1.625$ Solve for $\sin(C)$: $1.95 \sin(C) = -1.625 + 0.65 = -0.975$ $\sin(C) = \frac{-0.975}{1.95} = -0.5$ Thus, $C = -\frac{\pi}{6}$ (since $\sin(-\frac{\pi}{6}) = -0.5$).

Final equation:

The height $h_m$ of the particle as a function of time $t$ is:

$h_m = 1.95 \sin\left(\frac{3}{2}t - \frac{\pi}{6}\right) - 0.65$

Would you like further clarification on any steps? Here are some related questions to deepen your understanding:

1. How does the amplitude of the sine function affect the motion of the particle?
2. Why is the vertical shift $D$ negative in this context?
3. How would the equation change if the period were different?
4. What role does the phase shift $C$ play in determining the starting point of the motion?
5. Can you derive the equation for the particle's motion using a cosine function instead of a sine function?

Tip:

The period of any sinusoidal function is inversely related to the frequency. If you know the total time for a complete cycle, you can directly calculate the value of $B$.