The linear density of a string is `1.9 xx 10^(-4)"kg/m."` A transverse wave on the string is described by the equation
`y=(0.021 m) sin[(2.0 m^(-1))x +(30 s^(-1))t].` What are (a) the wave speed and (b) the tension in the string?

Answer

To solve the given problem, we will follow these steps:

### Given Data:
- Linear density of the string, \( \mu = 1.9 \times 10^{-4} \, \text{kg/m} \)
- Wave equation: \( y = (0.021 \, \text{m}) \sin[(2.0 \, \text{m}^{-1})x + (30 \, \text{s}^{-1})t] \)

### Step 1: Identify the angular frequency (\( \omega \)) and the wave number (\( k \))
From the wave equation, we can identify:
- The coefficient of \( t \) gives us the angular frequency:
\[ \omega = 30 \, \text{s}^{-1} \]
- The coefficient of \( x \) gives us the wave number:
\[ k = 2.0 \, \text{m}^{-1} \]

### Step 2: Calculate the wave speed (\( v \))
The relationship between angular frequency, wave speed, and wave number is given by:
\[ \omega = v k \]
Rearranging this to solve for \( v \):
\[ v = \frac{\omega}{k} \]
Substituting the values we found:
\[ v = \frac{30 \, \text{s}^{-1}}{2.0 \, \text{m}^{-1}} = 15 \, \text{m/s} \]

### Step 3: Calculate the tension in the string (\( T \))
The wave speed is also related to the tension and linear density by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\mu} \]
Rearranging to solve for \( T \):
\[ T = v^2 \mu \]
Substituting the values:
\[ T = (15 \, \text{m/s})^2 \times (1.9 \times 10^{-4} \, \text{kg/m}) = 225 \times 1.9 \times 10^{-4} \]
Calculating this gives:
\[ T = 0.04275 \, \text{N} \approx 0.043 \, \text{N} \]

### Final Answers:
(a) Wave speed, \( v = 15 \, \text{m/s} \)
(b) Tension in the string, \( T \approx 0.043 \, \text{N} \)

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