A string of length L=0.800 m is fixed at both ends. It oscillates in a four-loop standing wave pattern at a frequency f, = 560 Hz . The mass of the string is m = 2.39 g. 1. a. What is the wavelength of the four-loop pattern (the 4th harmonic), 14 , for this string? b. What is the wave speed, v, on the string? c. What is the tension, t for this string?
A string of length L=0.800 m is fixed at both ends. It oscillates in a four-loop standing wave pattern at a frequency f, = 560 Hz . The mass of the string is m = 2.39 g. 1. a. What is the wavelength of the four-loop pattern (the 4th harmonic), 14 , for this string? b. What is the wave speed, v, on the string? c. What is the tension, t for this string?
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![### Problem 1: Oscillations of a String
A string of length \( L = 0.800 \, \text{m} \) is fixed at both ends. It oscillates in a four-loop standing wave pattern at a frequency \( f_4 = 560 \, \text{Hz} \). The mass of the string is \( m = 2.39 \, \text{g} \).
**a. What is the wavelength of the four-loop pattern (the 4th harmonic), \( \lambda_4 \), for this string?**
For a string fixed at both ends and oscillating in its nth harmonic, the wavelength \(\lambda_n\) is given by:
\[ \lambda_n = \frac{2L}{n} \]
For the 4th harmonic (n = 4):
\[ \lambda_4 = \frac{2 \times 0.800 \, \text{m}}{4} = 0.400 \, \text{m} \]
**b. What is the wave speed, \( v \), on the string?**
The wave speed \( v \) can be calculated using the relationship:
\[ v = f \lambda \]
For the 4th harmonic:
\[ v = 560 \, \text{Hz} \times 0.400 \, \text{m} = 224 \, \text{m/s} \]
**c. What is the tension, \( \tau \), for this string?**
The tension \( \tau \) in the string is related to the wave speed \( v \), the mass \( m \), and the length \( L \) of the string by the formula:
\[ v = \sqrt{\frac{\tau}{\mu}} \]
where \(\mu\) is the linear mass density, given by:
\[ \mu = \frac{m}{L} \]
First, convert mass to kilograms:
\[ m = 2.39 \, \text{g} = 0.00239 \, \text{kg} \]
Thus:
\[ \mu = \frac{0.00239 \, \text{kg}}{0.800 \, \text{m}} = 0.0029875 \, \text{kg/m} \]
Now solve for the tension:
\[ v^2 = \frac{\tau}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb067f6a6-d6d0-4a81-90c2-26a75bd86046%2F48311857-71d4-46b9-b2ae-ab868bda90c8%2Fz1rr09_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 1: Oscillations of a String
A string of length \( L = 0.800 \, \text{m} \) is fixed at both ends. It oscillates in a four-loop standing wave pattern at a frequency \( f_4 = 560 \, \text{Hz} \). The mass of the string is \( m = 2.39 \, \text{g} \).
**a. What is the wavelength of the four-loop pattern (the 4th harmonic), \( \lambda_4 \), for this string?**
For a string fixed at both ends and oscillating in its nth harmonic, the wavelength \(\lambda_n\) is given by:
\[ \lambda_n = \frac{2L}{n} \]
For the 4th harmonic (n = 4):
\[ \lambda_4 = \frac{2 \times 0.800 \, \text{m}}{4} = 0.400 \, \text{m} \]
**b. What is the wave speed, \( v \), on the string?**
The wave speed \( v \) can be calculated using the relationship:
\[ v = f \lambda \]
For the 4th harmonic:
\[ v = 560 \, \text{Hz} \times 0.400 \, \text{m} = 224 \, \text{m/s} \]
**c. What is the tension, \( \tau \), for this string?**
The tension \( \tau \) in the string is related to the wave speed \( v \), the mass \( m \), and the length \( L \) of the string by the formula:
\[ v = \sqrt{\frac{\tau}{\mu}} \]
where \(\mu\) is the linear mass density, given by:
\[ \mu = \frac{m}{L} \]
First, convert mass to kilograms:
\[ m = 2.39 \, \text{g} = 0.00239 \, \text{kg} \]
Thus:
\[ \mu = \frac{0.00239 \, \text{kg}}{0.800 \, \text{m}} = 0.0029875 \, \text{kg/m} \]
Now solve for the tension:
\[ v^2 = \frac{\tau}{
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