A transversee wave propagating on the string can be described by the equation` y= 2 sin (10 x + 300 t) ` , where x and y are in metres and t in second. If the vibrating string has linear density of ` 0.6 xx 10^(-3) g//cm` , then the tension in the string is

Options

A. 5.4N
B. 0.054 N
C. 54 N
D. 0.0054 N

Answer

To find the tension in the string given the wave equation and the linear density, we can follow these steps:

### Step 1: Identify the wave parameters from the wave equation
The wave equation is given as:
\[ y = 2 \sin(10x + 300t) \]
From this equation, we can identify:
- Amplitude \( A = 2 \) m
- Wave number \( k = 10 \) m\(^{-1}\)
- Angular frequency \( \omega = 300 \) rad/s

### Step 2: Calculate the velocity of the wave
The velocity \( v \) of a wave can be calculated using the relationship between wave number \( k \) and angular frequency \( \omega \):
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{300}{10} = 30 \text{ m/s} \]

### Step 3: Convert the linear density to the correct units
The linear density \( \mu \) is given as:
\[ \mu = 0.6 \times 10^{-3} \text{ g/cm} \]
To convert this to kg/m, we perform the following conversions:
1. Convert grams to kilograms: \( 0.6 \times 10^{-3} \text{ g} = 0.6 \times 10^{-6} \text{ kg} \)
2. Convert cm to m: \( 1 \text{ cm} = 0.01 \text{ m} \)
Thus,
\[ \mu = 0.6 \times 10^{-3} \text{ g/cm} = 0.6 \times 10^{-3} \text{ g} \times \frac{100}{1} \text{ cm/m} = 0.6 \times 10^{-1} \text{ kg/m} = 0.6 \times 10^{-2} \text{ kg/m} = 6 \times 10^{-4} \text{ kg/m} \]

### Step 4: Calculate the tension in the string
The tension \( T \) in the string can be calculated using the formula:
\[ T = \mu v^2 \]
Substituting the values we have:
\[ T = (6 \times 10^{-4} \text{ kg/m}) \times (30 \text{ m/s})^2 \]
Calculating \( 30^2 \):
\[ 30^2 = 900 \]
Now substituting back:
\[ T = (6 \times 10^{-4}) \times 900 = 540 \times 10^{-4} \text{ N} = 5.4 \times 10^{-2} \text{ N} \]

### Final Answer
The tension in the string is:
\[ T = 0.054 \text{ N} \]

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