A thin string is held at one end and oscillates so that,
`y(x = 0, t) = 8 sin 4t(cm)`
Neglect the gravitattional force. The dtring's linear mass density is `0 .2 kg// m` and its tension is ` 1 N`. The string passes through a bath filled with `1 kg` water. Due to friction heat is transferred to the bath. The heat transfer efficiency is `50%`. Calculate how much time passes before the temperature of the bath rises one degree kelvin?

Explanation

To solve the problem step by step, we will follow these main steps:

### Step 1: Identify the parameters from the wave equation
The wave equation given is:
\[ y(x = 0, t) = 8 \sin(4t) \text{ cm} \]
From this, we can identify:
- Amplitude \( A = 8 \) cm = 0.08 m (convert to meters)
- Angular frequency \( \omega = 4 \) rad/s

### Step 2: Calculate the linear mass density and tension
The linear mass density \( \mu \) is given as:
\[ \mu = 0.2 \text{ kg/m} \]
The tension \( T \) in the string is given as:
\[ T = 1 \text{ N} \]

### Step 3: Calculate the wave speed
The speed of the wave \( v \) on the string can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{1 \text{ N}}{0.2 \text{ kg/m}}} = \sqrt{5} \approx 2.24 \text{ m/s} \]

### Step 4: Calculate the average power of the wave
The average power \( P \) transmitted by the wave is given by:
\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]
Substituting the values:
- \( A = 0.08 \) m
- \( \omega = 4 \) rad/s
- \( v \approx 2.24 \) m/s

Calculating:
\[ P = \frac{1}{2} \times 0.2 \times (0.08)^2 \times (4)^2 \times 2.24 \]
\[ P = \frac{1}{2} \times 0.2 \times 0.0064 \times 16 \times 2.24 \]
\[ P = 0.5 \times 0.2 \times 0.0064 \times 16 \times 2.24 \approx 0.000229 \text{ W} \]

### Step 5: Calculate the effective power transferred to the bath
Since the heat transfer efficiency is 50%, the effective power \( P' \) transferred to the bath is:
\[ P' = 0.5 \times P \]
\[ P' = 0.5 \times 0.000229 \approx 0.0001145 \text{ W} \]

### Step 6: Calculate the energy required to raise the temperature of the water
The energy \( Q \) required to raise the temperature of the water by 1 K is given by:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m = 1 \text{ kg} \) (mass of water)
- \( c = 4.2 \times 10^3 \text{ J/(kg K)} \) (specific heat of water)
- \( \Delta T = 1 \text{ K} \)

Calculating:
\[ Q = 1 \cdot 4.2 \times 10^3 \cdot 1 = 4200 \text{ J} \]

### Step 7: Calculate the time required to raise the temperature
Using the formula:
\[ Q = P' \cdot t \]
We can rearrange this to find time \( t \):
\[ t = \frac{Q}{P'} \]
Substituting the values:
\[ t = \frac{4200}{0.0001145} \approx 3.67 \times 10^5 \text{ seconds} \]

### Step 8: Convert time to days
To convert seconds to days:
\[ t \approx \frac{3.67 \times 10^5}{86400} \approx 4.25 \text{ days} \]

### Final Answer
The time that passes before the temperature of the bath rises by one degree Kelvin is approximately **4.25 days**.

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