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Answer :
To find the fundamental frequency of a stretched string with both ends fixed, we can use the following formula:
[tex]\[ f = \frac{1}{2L} \times \sqrt{\frac{T}{\mu}} \][/tex]
Where:
- [tex]\( f \)[/tex] is the fundamental frequency.
- [tex]\( L \)[/tex] is the length of the string in meters.
- [tex]\( T \)[/tex] is the tension in the string, in newtons (N).
- [tex]\( \mu \)[/tex] is the mass per unit length of the string, in kg/m.
Let's go through the steps to solve the problem:
1. Convert the length of the string to meters:
- Given length is 181 cm.
- Convert to meters: [tex]\( L = 181 \, \text{cm} \times 0.01 \, \frac{\text{m}}{\text{cm}} = 1.81 \, \text{m} \)[/tex]
2. Convert the density to kg/m:
- Given density is 0.038 g/cm.
- Convert to kg/m (since 1 g/cm = 10 kg/m): [tex]\( \text{Density in kg/m} = 0.038 \, \text{g/cm} \times 10 = 0.38 \, \text{kg/m} \)[/tex]
3. Calculate the mass per unit length ([tex]\(\mu\)[/tex]):
- Since the density is already in kg/m, and we need it in terms of kg/m for our formula, we can treat it as [tex]\(\mu = 0.38 \, \text{kg/m}\)[/tex].
4. Given the tension [tex]\(T\)[/tex] is 1650 N.
5. Calculate the fundamental frequency ([tex]\(f\)[/tex]):
- Use the formula:
[tex]\[ f = \frac{1}{2 \times 1.81} \times \sqrt{\frac{1650}{0.38}} \][/tex]
- [tex]\(\sqrt{\frac{1650}{0.38}} \approx 44.314\)[/tex]
- [tex]\(\frac{1}{2 \times 1.81} \approx 0.276\)[/tex]
- So, [tex]\( f \approx 0.276 \times 44.314 \approx 13.53 \, \text{Hz} \)[/tex]
Therefore, the fundamental frequency of the string is approximately [tex]\(13.53 \, \text{Hz}\)[/tex].
[tex]\[ f = \frac{1}{2L} \times \sqrt{\frac{T}{\mu}} \][/tex]
Where:
- [tex]\( f \)[/tex] is the fundamental frequency.
- [tex]\( L \)[/tex] is the length of the string in meters.
- [tex]\( T \)[/tex] is the tension in the string, in newtons (N).
- [tex]\( \mu \)[/tex] is the mass per unit length of the string, in kg/m.
Let's go through the steps to solve the problem:
1. Convert the length of the string to meters:
- Given length is 181 cm.
- Convert to meters: [tex]\( L = 181 \, \text{cm} \times 0.01 \, \frac{\text{m}}{\text{cm}} = 1.81 \, \text{m} \)[/tex]
2. Convert the density to kg/m:
- Given density is 0.038 g/cm.
- Convert to kg/m (since 1 g/cm = 10 kg/m): [tex]\( \text{Density in kg/m} = 0.038 \, \text{g/cm} \times 10 = 0.38 \, \text{kg/m} \)[/tex]
3. Calculate the mass per unit length ([tex]\(\mu\)[/tex]):
- Since the density is already in kg/m, and we need it in terms of kg/m for our formula, we can treat it as [tex]\(\mu = 0.38 \, \text{kg/m}\)[/tex].
4. Given the tension [tex]\(T\)[/tex] is 1650 N.
5. Calculate the fundamental frequency ([tex]\(f\)[/tex]):
- Use the formula:
[tex]\[ f = \frac{1}{2 \times 1.81} \times \sqrt{\frac{1650}{0.38}} \][/tex]
- [tex]\(\sqrt{\frac{1650}{0.38}} \approx 44.314\)[/tex]
- [tex]\(\frac{1}{2 \times 1.81} \approx 0.276\)[/tex]
- So, [tex]\( f \approx 0.276 \times 44.314 \approx 13.53 \, \text{Hz} \)[/tex]
Therefore, the fundamental frequency of the string is approximately [tex]\(13.53 \, \text{Hz}\)[/tex].
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