The strings of a violin are tuned to the tones `G , D , A and E` which are separated by a fifth from one another . That is `f(D) = 1.5 (G) , f(A) = 1.5 f(D) = 400 Hz and f(E ) = 1.5 f(A)`. The distance between the two fixed points , the bridge at the scroll and over the body of the instrument is `0.25 m`. The tension on the string `E is 90 N`. The mass per unit length of string `E` is nearly

Options

A. ` 1 g//m`
B. ` 2 g//m`
C. ` 3 g// m`
D. `4 g//m`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the frequency of string E
We know from the problem that:
- \( f(D) = 1.5 \times f(G) \)
- \( f(A) = 1.5 \times f(D) = 400 \, \text{Hz} \)
- \( f(E) = 1.5 \times f(A) \)

First, we calculate \( f(E) \):
\[ f(E) = 1.5 \times 400 \, \text{Hz} = 600 \, \text{Hz} \]

### Step 2: Use the formula for frequency of a vibrating string
The frequency of a vibrating string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f \) is the frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 3: Substitute the known values into the formula
We know:
- \( f(E) = 600 \, \text{Hz} \)
- \( L = 0.25 \, \text{m} \)
- \( T = 90 \, \text{N} \)

Substituting these values into the frequency formula:
\[ 600 = \frac{1}{2 \times 0.25} \sqrt{\frac{90}{\mu}} \]

### Step 4: Simplify the equation
First, simplify \( \frac{1}{2 \times 0.25} \):
\[ \frac{1}{2 \times 0.25} = \frac{1}{0.5} = 2 \]
So the equation becomes:
\[ 600 = 2 \sqrt{\frac{90}{\mu}} \]

### Step 5: Isolate the square root term
Divide both sides by 2:
\[ 300 = \sqrt{\frac{90}{\mu}} \]

### Step 6: Square both sides to eliminate the square root
\[ 300^2 = \frac{90}{\mu} \]
\[ 90000 = \frac{90}{\mu} \]

### Step 7: Solve for \( \mu \)
Rearranging gives:
\[ \mu = \frac{90}{90000} \]
\[ \mu = \frac{1}{1000} \, \text{kg/m} = 0.001 \, \text{kg/m} \]

### Step 8: Convert to grams per meter
Since \( 1 \, \text{kg} = 1000 \, \text{grams} \):
\[ \mu = 0.001 \, \text{kg/m} \times 1000 \, \text{g/kg} = 1 \, \text{g/m} \]

Thus, the mass per unit length of string E is nearly **1 gram per meter**.

### Final Answer
The mass per unit length of string E is approximately **1 gram per meter**.

---