Let's work through the given problem step-by-step to construct the graph and find the appropriate function.

Part A: Sketching the Graph

1. Understanding the Situation:

   • The equilibrium position of the object is 25 cm above the table.
   • The object oscillates 12 cm above and below this equilibrium position, so:
     • Maximum Height: $25 + 12 = 37$ cm.
     • Minimum Height: $25 - 12 = 13$ cm.
   • The midline (equilibrium position) is at 25 cm.

2. Time Analysis:

   • At $t = 0$, the object is at its maximum height of 37 cm.
   • It reaches its minimum height 0.3 seconds after reaching the maximum.
   • This means it took 0.3 seconds to travel from the maximum to the minimum, i.e., half of a complete cycle.
   • Therefore, a complete cycle (period) is twice this time, or $2 \times 0.3 = 0.6$ seconds.

3. Key Points for the Graph:

   • Maximum: At $t = 0$, height is 37 cm.
   • Minimum: At $t = 0.3$, height is 13 cm.
   • One complete cycle ends at $t = 0.6$, bringing the height back to the maximum of 37 cm.

4. Graph Components:

   • X-Axis (Time): Label it with a scale from $t = 0$ to $t = 0.6$ seconds (or beyond for additional cycles).
   • Y-Axis (Height): Mark the maximum, minimum, and midline (13, 25, and 37 cm).
   • Plot the sinusoidal wave showing a complete cycle.

Part B: Formulating the Equation

We want a function of the form: $h(t) = a \cos(b(t - c)) + d$ where:

• $a$ = Amplitude.
• $b$ = Frequency factor.
• $c$ = Horizontal shift.
• $d$ = Vertical shift.

1. Identify the Constants:

   • Amplitude $a$: This is the distance from the midline to the maximum (or minimum). $a = 12$.
   • Vertical Shift $d$: This is the midline of the oscillation, which is 25 cm. So, $d = 25$.
   • Frequency Factor $b$: The period of the cosine function is given by $\frac{2\pi}{b}$. We know the period is 0.6 seconds, so: $\frac{2\pi}{b} = 0.6 \implies b = \frac{2\pi}{0.6} = \frac{10\pi}{3}$
   • Horizontal Shift $c$: Since the object starts at the maximum height at $t = 0$, there is no phase shift, i.e., $c = 0$.

2. Constructing the Function: Putting it all together: $h(t) = 12 \cos \left( \frac{10\pi}{3} t \right) + 25$

3. Finding the Height at $t = 3.4$ Seconds: Substitute $t = 3.4$ into the function: $h(3.4) = 12 \cos \left( \frac{10\pi}{3} \times 3.4 \right) + 25$

I will calculate this value to determine the exact height.The height of the object 3.4 seconds after the stopwatch started is approximately 19 cm.

Summary:

1. Equation:
   $h(t) = 12 \cos \left( \frac{10\pi}{3} t \right) + 25$

2. Height at $t = 3.4$ seconds:
   $h(3.4) \approx 19$ cm.

Would you like to see a detailed breakdown of any step, or do you have any specific questions?

Related Questions:

1. What would be the object's height at $t = 1$ second?
2. How would the graph change if the amplitude increased to 15 cm?
3. What is the object's velocity at $t = 0.3$ seconds?
4. How would the period change if the frequency factor $b$ was doubled?
5. What is the displacement function for this oscillating object?

Tip: When analyzing oscillatory motion, always identify the amplitude, midline, and period first for better clarity in constructing equations.