Answer :
To find the tension required for the vibrating wire to produce the correct frequency of 4100 Hz (4.1 kHz), we can follow these steps:
### Step 1: Identify Given Values
- Length of the wire ([tex]\(L\)[/tex]): [tex]\(0.050 \text{ m}\)[/tex]
- Diameter of the wire ([tex]\(d\)[/tex]): [tex]\(3.5 \times 10^{-4} \text{ m}\)[/tex]
- Density of the wire ([tex]\(\rho\)[/tex]): [tex]\(7.8 \times 10^3 \text{ kg/m}^3\)[/tex]
- Frequency ([tex]\(f\)[/tex]): [tex]\(4100 \text{ Hz}\)[/tex]
### Step 2: Calculate Cross-Sectional Area
The cross-sectional area [tex]\(A\)[/tex] of the wire can be calculated using the formula for the area of a circle:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Plugging in the given diameter:
[tex]\[ A = \pi \left( \frac{3.5 \times 10^{-4}}{2} \right)^2 \approx 9.62 \times 10^{-8} \text{ m}^2 \][/tex]
### Step 3: Calculate Mass Per Unit Length
The mass per unit length [tex]\(\mu\)[/tex] of the wire is given by:
[tex]\[ \mu = \rho \times A \][/tex]
Substituting the density and the cross-sectional area:
[tex]\[ \mu = 7.8 \times 10^3 \text{ kg/m}^3 \times 9.62 \times 10^{-8} \text{ m}^2 \approx 0.00075 \text{ kg/m} \][/tex]
### Step 4: Calculate the Tension
The tension [tex]\(T\)[/tex] in the wire can be found using the formula for the fundamental frequency of a vibrating string:
[tex]\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \][/tex]
Solving for [tex]\(T\)[/tex]:
[tex]\[ T = \left( 2Lf \right)^2 \mu \][/tex]
Substituting the given values:
[tex]\[ T = \left( 2 \times 0.050 \text{ m} \times 4100 \text{ Hz} \right)^2 \times 0.00075 \text{ kg/m} \][/tex]
[tex]\[ T = \left( 410 \text{ s}^{-1} \right)^2 \times 0.00075 \text{ kg/m} \][/tex]
[tex]\[ T = 168100 \times 0.00075 \text{ N} \][/tex]
[tex]\[ T \approx 126.15 \text{ N} \][/tex]
Thus, the tension required for the vibrating wire to produce its correct frequency of 4100 Hz is approximately [tex]\(126.15 \text{ N}\)[/tex].
### Step 1: Identify Given Values
- Length of the wire ([tex]\(L\)[/tex]): [tex]\(0.050 \text{ m}\)[/tex]
- Diameter of the wire ([tex]\(d\)[/tex]): [tex]\(3.5 \times 10^{-4} \text{ m}\)[/tex]
- Density of the wire ([tex]\(\rho\)[/tex]): [tex]\(7.8 \times 10^3 \text{ kg/m}^3\)[/tex]
- Frequency ([tex]\(f\)[/tex]): [tex]\(4100 \text{ Hz}\)[/tex]
### Step 2: Calculate Cross-Sectional Area
The cross-sectional area [tex]\(A\)[/tex] of the wire can be calculated using the formula for the area of a circle:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Plugging in the given diameter:
[tex]\[ A = \pi \left( \frac{3.5 \times 10^{-4}}{2} \right)^2 \approx 9.62 \times 10^{-8} \text{ m}^2 \][/tex]
### Step 3: Calculate Mass Per Unit Length
The mass per unit length [tex]\(\mu\)[/tex] of the wire is given by:
[tex]\[ \mu = \rho \times A \][/tex]
Substituting the density and the cross-sectional area:
[tex]\[ \mu = 7.8 \times 10^3 \text{ kg/m}^3 \times 9.62 \times 10^{-8} \text{ m}^2 \approx 0.00075 \text{ kg/m} \][/tex]
### Step 4: Calculate the Tension
The tension [tex]\(T\)[/tex] in the wire can be found using the formula for the fundamental frequency of a vibrating string:
[tex]\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \][/tex]
Solving for [tex]\(T\)[/tex]:
[tex]\[ T = \left( 2Lf \right)^2 \mu \][/tex]
Substituting the given values:
[tex]\[ T = \left( 2 \times 0.050 \text{ m} \times 4100 \text{ Hz} \right)^2 \times 0.00075 \text{ kg/m} \][/tex]
[tex]\[ T = \left( 410 \text{ s}^{-1} \right)^2 \times 0.00075 \text{ kg/m} \][/tex]
[tex]\[ T = 168100 \times 0.00075 \text{ N} \][/tex]
[tex]\[ T \approx 126.15 \text{ N} \][/tex]
Thus, the tension required for the vibrating wire to produce its correct frequency of 4100 Hz is approximately [tex]\(126.15 \text{ N}\)[/tex].
