give answer in 2 step with explanation at the end of each step and final answer at the end:Activity 16B-1
A rubber ducky (located atx = 0m) floatson ~~ o,1 "
a lake and bobs up and down due to a wat

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Activity 16B-1
A rubber ducky (located atx = 0m) floatson ~~ o,1 "
a lake and bobs up and down due to a water,
wave. Shown below are a history graph,
(0), that depicts the vertical displacement of 7 7 T Tw
rubber ducky atx = 0 m as a function of a1
time and a snapshot graph, (x), that depicts ©
the water wave as a function of position at | 470941°¢
1=0s. A
a. Explain how these graphs tell you «
‘whether this wave is moving to the o 0 ? xn)
right or left a
b. What is the magnitude and direction of the velocity of this wave?
©. Write out the equation of motion y(x,r) = tsn{2x 2x £9) for this wave.

Make sure you specify the values for 4, T, 4, , =, and D.
a
Activity 16B-2
A speaker creates uniformly spherical sound with 750 Watts of

power.

a) What is the intensity, I, of the sound at a distance of 100
meters from the speaker? What is the sound intensity level, 8,
of the sound at a distance of 100 meters from the speaker?

b) What is the intensity, I, of the sound at a distance of 50
meters from the speaker? What is the sound intensity level, 8,
of the sound at a di

VIPSolutions · Accepted Answer

Let's break down each part in **2 steps** with explanations after each step and final answers at the end.

---

## Activity 16B-1

### a. Wave Direction

**Step 1:**  
Compare the *history graph* $ y(t) $ at $ x= $ and the *snapshot graph* $ y(x) $ at $ t= $.  
- Notice where the peaks/troughs occur in both graphs.
- The *history graph* shows motion over time at a fixed point; the *snapshot graph* shows motion over space at a fixed instant.

**Explanation:**  
If, as time increases, the wave at $ x= $ matches what you see at greater $ x $ in the snapshot, the wave is moving in the **positive x-direction (right)**. If it matches smaller $ x $, it's moving **left**.

**Step 2:**  
From the graphs, at $ t= $, the peak is at $ x= $ in the snapshot. At $ t> $, the peak at $ x= $ drops, and the peak appears at increasing $ x $ in the snapshot.

**Explanation:**  
This means the wave is moving to the **right**.

**Final Answer:**  
The wave is **moving to the right (positive x-direction)**.

---

### b. Magnitude and Direction of Velocity

**Step 1:**  
Find the **wavelength ($ \lambda $)** and **period ($ T $)** from the graphs:
- Wavelength ($ \lambda $) is the distance between two peaks in the snapshot graph ($ y(x) $).
- Period ($ T $) is the time between two peaks in the history graph ($ y(t) $).

From the graphs:
- $ \lambda = 2\, \text{m} $ (distance between peaks in position graph)
- $ T = 4\, \text{s} $ (time between peaks in time graph)

**Explanation:**  
The wave speed is $ v = \frac{\lambda}{T} $.

**Step 2:**  
Calculate the speed:
$$
v = \frac{2\, \text{m}}{4\, \text{s}} = .5\, \text{m/s}
$$
Direction is **to the right** (from part a).

**Explanation:**  
The positive value matches the direction found in (a).

**Final Answer:**  
**Magnitude:** $ .5\, \text{m/s} $  
**Direction:** To the right

---

### c. Wave Equation

**Step 1:**  
Write the general form:
$$
y(x, t) = A \sin\left(2\pi \frac{x}{\lambda} - 2\pi \frac{t}{T} + \phi \right)
$$
From graphs:
- Amplitude ($ A $): $ .2\, \text{m} $
- Wavelength ($ \lambda $): $ 2\, \text{m} $
- Period ($ T $): $ 4\, \text{s} $
- Phase ($ \phi $): At $ x=, t= $, $ y= $, so $ \phi= $
- Direction: minus sign since wave moves right

**Explanation:**  
Plug in the values from the graphs.

**Step 2:**  
$$
y(x, t) = .2 \sin\left(2\pi \frac{x}{2} - 2\pi \frac{t}{4}\right)
= .2 \sin\left(\pi x - \frac{\pi}{2} t\right)
$$

**Explanation:**  
All values substituted as required.

**Final Answer:**  
$$
y(x, t) = .2 \sin\left(\pi x - \frac{\pi}{2} t\right)
$$

---

## Activity 16B-2

### a. Intensity and Sound Intensity Level at 100 m

**Step 1:**  
Intensity, $ I = \frac{P}{4\pi r^2} $  
- $ P = 750\, \text{W} $
- $ r = 100\, \text{m} $

$$
I = \frac{750}{4\pi (100)^2} = \frac{750}{125,663.7} \approx .006 \text{ W/m}^2
$$

**Explanation:**  
Intensity is power per unit area for a sphere.

**Step 2:**  
Sound intensity level, $ \beta = 10 \log_{10} \left( \frac{I}{I_} \right) $  
- $ I_ = 1 \times 10^{-12}\, \text{W/m}^2 $

$$
\beta = 10 \log_{10} \left( \frac{.006}{1\times10^{-12}} \right) \approx 10 \times 9.78 = 97.8 \text{ dB}
$$

**Explanation:**  
Convert intensity to dB using the reference level.

**Final Answer:**  
- **Intensity at 100 m:** $ .006\, \text{W/m}^2 $
- **Sound intensity level at 100 m:** $ 97.8\, \text{dB} $

---

### b. Intensity and Sound Intensity Level at 50 m

**Step 1:**  
$$
I = \frac{750}{4\pi (50)^2} = \frac{750}{31,415.9} \approx .024 \text{ W/m}^2
$$

**Explanation:**  
Smaller distance means higher intensity.

**Step 2:**  
$$
\beta = 10 \log_{10} \left( \frac{.024}{1\times10^{-12}} \right) \approx 10 \times 10.38 = 103.8 \text{ dB}
$$

**Explanation:**  
Calculate dB level as before.

**Final Answer:**  
- **Intensity at 50 m:** $ .024\, \text{W/m}^2 $
- **Sound intensity level at 50 m:** $ 103.8\, \text{dB} $

---

### c. Increase in Decibels from 100 m to 50 m

**Step 1:**  
Subtract the dB at 100 m from the dB at 50 m:
$$
103.8\, \text{dB} - 97.8\, \text{dB} = 6\, \text{dB}
$$

**Explanation:**  
Difference in dB levels gives the increase.

**Step 2:**  
The increase is **6 dB**.

**Explanation:**  
Every time you halve the distance, intensity quadruples, which increases dB by 6.

**Final Answer:**  
**Increase:** $ 6\, \text{dB} $

---

### d. Power Needed for 100 dB at 50 m

**Step 1:**  
First, find the required intensity for 100 dB:
$$
100 = 10 \log_{10} \left( \frac{I}{1\times10^{-12}} \right) \implies I = 1\times10^{-10} \text{ W/m}^2
$$

**Explanation:**  
Solve for $ I $ using the dB formula.

**Step 2:**  
The area at 50 m: $ A = 4\pi (50)^2 = 31,415.9\, \text{m}^2 $.  
Required power: $ P = I \times A = 1\times10^{-10} \times 31,415.9 = 3.14 \times 10^{-6} \text{ W} $.

**Explanation:**  
Multiply the intensity by the area to get power.

**Final Answer:**  
**Required power output:** $ 3.14 \times 10^{-6} \text{ W} $ (or about *

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