Answer :
To find the tension required for a violin string to vibrate at a fundamental frequency of 440 Hz, we use the properties of the string and the formula for wave frequency on a stretched string. Here's a step-by-step guide to solving it:
1. Understand the Variables:
- Frequency (f): 440 Hz (this is the sound frequency the string produces).
- Length (L) of the string: 0.33 m.
- Diameter of the string: 0.05 cm.
- Density (ρ) of the material: 3.5 x 10³ kg/m³.
2. Convert Units Where Necessary:
- Convert the diameter from centimeters to meters:
[tex]\[
\text{Diameter in meters} = 0.05 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.0005 \, \text{m}
\][/tex]
3. Calculate the Cross-Sectional Area (A):
- The string is cylindrical, so its cross-sectional area is the area of a circle:
[tex]\[
A = \pi \times \left(\frac{\text{Diameter}}{2}\right)^2 = \pi \times (0.00025)^2 \approx 1.963 \times 10^{-7} \, \text{m}^2
\][/tex]
4. Calculate the Linear Mass Density (μ):
- This is the mass per unit length of the string:
[tex]\[
\mu = \text{Density} \times \text{Cross-Sectional Area} = 3.5 \times 10^3 \, \text{kg/m}^3 \times 1.963 \times 10^{-7} \, \text{m}^2 \approx 0.000687 \, \text{kg/m}
\][/tex]
5. Use the Formula for the Fundamental Frequency:
- The frequency of a vibrating string is given by:
[tex]\[
f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\][/tex]
- Rearranging to solve for the tension (T):
[tex]\[
T = (2Lf)^2 \times \mu
\][/tex]
6. Calculate the Tension (T):
- Substitute in the values we know:
[tex]\[
T = (2 \times 0.33 \times 440)^2 \times 0.000687
\][/tex]
- Simplifying:
[tex]\[
= 57.955 \, \text{Newtons}
\][/tex]
Therefore, the tension required in the violin string for it to vibrate at the fundamental frequency of 440 Hz is approximately 57.96 Newtons.
1. Understand the Variables:
- Frequency (f): 440 Hz (this is the sound frequency the string produces).
- Length (L) of the string: 0.33 m.
- Diameter of the string: 0.05 cm.
- Density (ρ) of the material: 3.5 x 10³ kg/m³.
2. Convert Units Where Necessary:
- Convert the diameter from centimeters to meters:
[tex]\[
\text{Diameter in meters} = 0.05 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.0005 \, \text{m}
\][/tex]
3. Calculate the Cross-Sectional Area (A):
- The string is cylindrical, so its cross-sectional area is the area of a circle:
[tex]\[
A = \pi \times \left(\frac{\text{Diameter}}{2}\right)^2 = \pi \times (0.00025)^2 \approx 1.963 \times 10^{-7} \, \text{m}^2
\][/tex]
4. Calculate the Linear Mass Density (μ):
- This is the mass per unit length of the string:
[tex]\[
\mu = \text{Density} \times \text{Cross-Sectional Area} = 3.5 \times 10^3 \, \text{kg/m}^3 \times 1.963 \times 10^{-7} \, \text{m}^2 \approx 0.000687 \, \text{kg/m}
\][/tex]
5. Use the Formula for the Fundamental Frequency:
- The frequency of a vibrating string is given by:
[tex]\[
f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\][/tex]
- Rearranging to solve for the tension (T):
[tex]\[
T = (2Lf)^2 \times \mu
\][/tex]
6. Calculate the Tension (T):
- Substitute in the values we know:
[tex]\[
T = (2 \times 0.33 \times 440)^2 \times 0.000687
\][/tex]
- Simplifying:
[tex]\[
= 57.955 \, \text{Newtons}
\][/tex]
Therefore, the tension required in the violin string for it to vibrate at the fundamental frequency of 440 Hz is approximately 57.96 Newtons.
