The breaking stress of steel is `7.85 xx 10^(8) N//m^(2)` and density of steel is `7.7 xx 10^(3) kg//m^(3)`. The maximum frequency to which a string `1 m` long can be tuned is

Options

A. `15.8 Hz`
B. `158 Hz`
C. `47.4 Hz`
D. `474 Hz`

Explanation

To find the maximum frequency to which a string of length 1 meter can be tuned, we can use the formula for the fundamental frequency of a string fixed at both ends:

\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Where:
- \( f_0 \) is the fundamental frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the linear mass density of the string.

### Step 1: Determine the Linear Mass Density (\( \mu \))

The linear mass density (\( \mu \)) can be expressed as:

\[ \mu = \frac{m}{L} \]

Where \( m \) is the mass of the string. The mass can be calculated using the density (\( \rho \)) of the material and the volume (\( V \)) of the string:

\[ m = \rho \cdot V \]

For a string of length \( L \) and cross-sectional area \( A \):

\[ V = A \cdot L \]

Thus,

\[ m = \rho \cdot A \cdot L \]

Now substituting this into the equation for \( \mu \):

\[ \mu = \frac{\rho \cdot A \cdot L}{L} = \rho \cdot A \]

### Step 2: Determine the Tension (\( T \))

The tension in the string can be related to the breaking stress (\( \sigma \)) and the cross-sectional area (\( A \)):

\[ T = \sigma \cdot A \]

Given that the breaking stress of steel is \( 7.85 \times 10^8 \, \text{N/m}^2 \), we can write:

\[ T = 7.85 \times 10^8 \cdot A \]

### Step 3: Substitute into the Frequency Formula

Now substituting \( T \) and \( \mu \) into the frequency formula:

\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{2L} \sqrt{\frac{7.85 \times 10^8 \cdot A}{\rho \cdot A}} \]

Notice that \( A \) cancels out:

\[ f_0 = \frac{1}{2L} \sqrt{\frac{7.85 \times 10^8}{\rho}} \]

### Step 4: Substitute Values

Now substituting the values for \( L \) and \( \rho \):

- \( L = 1 \, \text{m} \)
- \( \rho = 7.7 \times 10^3 \, \text{kg/m}^3 \)

\[ f_0 = \frac{1}{2 \times 1} \sqrt{\frac{7.85 \times 10^8}{7.7 \times 10^3}} \]

Calculating the value inside the square root:

\[ \frac{7.85 \times 10^8}{7.7 \times 10^3} = 1.01818 \times 10^5 \]

Taking the square root:

\[ \sqrt{1.01818 \times 10^5} \approx 318.5 \]

Now substituting back into the frequency equation:

\[ f_0 = \frac{1}{2} \times 318.5 \approx 159.25 \, \text{Hz} \]

### Conclusion

Thus, the maximum frequency to which a string 1 meter long can be tuned is approximately:

\[ f_0 \approx 158 \, \text{Hz} \]

### Final Answer

The maximum frequency is \( 158 \, \text{Hz} \).

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