Answered

A piano repairer replaces the wire that produces the highest note on a piano. The wire has a vibrating length of 0.050 m. He uses a wire with the following properties:

- Diameter: [tex]\(3.5 \times 10^{-4} \, \text{m}\)[/tex]
- Density: [tex]\(7.8 \times 10^3 \, \text{kg} \, \text{m}^{-3}\)[/tex]
- Breaking stress: [tex]\(3.0 \times 10^9 \, \text{N} \, \text{m}^{-2}\)[/tex]

Calculate the tension required for the vibrating wire to produce its correct frequency of 4.1 kHz.



Answer :

GinnyAnswer
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].