College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem Statement:**
Yujin is playing on a violin string with a length of 32 cm and a linear density of 1.5 g/cm, that resonates with the first overtone from a 1.7-m long organ pipe with one end closed and the other end open that Jeong is playing. What is the tension in the string? Assume the speed of sound in the air is 343 m/s.
**Explanation:**
In this problem, we need to determine the tension in the violin string, which resonates with a particular organ pipe note. The steps involve converting units, calculating frequencies for both the string and the pipe, and then using the equations of waves on a string to find the tension.
1. **Unit Conversion and Parameters:**
- Length of violin string, \( L_s = 32 \text{ cm} = 0.32 \text{ m} \)
- Linear density of violin string, \( \mu = 1.5 \text{ g/cm} = 0.0015 \text{ kg/m} \)
- Length of organ pipe, \( L_p = 1.7 \text{ m} \)
- Speed of sound in air, \( v = 343 \text{ m/s} \)
2. **Organ Pipe:**
- For an organ pipe with one end closed and the other open, it supports odd harmonics.
- First overtone (second harmonic) frequency formula:
\[
f_1 = \frac{3v}{4L_p} = \frac{3 \times 343}{4 \times 1.7} \approx 151 \text{ Hz}
\]
3. **Violin String:**
- The string's frequency \( f_s \) matches the frequency of the organ pipe's first overtone.
- Fundamental frequency of a string:
\[
f_s = \frac{1}{2L_s}\sqrt{\frac{T}{\mu}}
\]
- Given \( f_s = 151 \text{ Hz} \), we rearrange the formula to solve for tension \( T \):
\[
151 = \frac{1}{2 \times 0.32} \sqrt{\frac{T}{0.0015}}
\]
\[
151 = \frac{1}{0.64} \](https://content.bartleby.com/qna-images/question/cb865a02-7408-4eb8-b9f3-2c56928a70f5/89fff3e2-72c8-45b6-b5b5-de0285de04d1/ssgiwgr_processed.png)
Transcribed Image Text:**Problem Statement:**
Yujin is playing on a violin string with a length of 32 cm and a linear density of 1.5 g/cm, that resonates with the first overtone from a 1.7-m long organ pipe with one end closed and the other end open that Jeong is playing. What is the tension in the string? Assume the speed of sound in the air is 343 m/s.
**Explanation:**
In this problem, we need to determine the tension in the violin string, which resonates with a particular organ pipe note. The steps involve converting units, calculating frequencies for both the string and the pipe, and then using the equations of waves on a string to find the tension.
1. **Unit Conversion and Parameters:**
- Length of violin string, \( L_s = 32 \text{ cm} = 0.32 \text{ m} \)
- Linear density of violin string, \( \mu = 1.5 \text{ g/cm} = 0.0015 \text{ kg/m} \)
- Length of organ pipe, \( L_p = 1.7 \text{ m} \)
- Speed of sound in air, \( v = 343 \text{ m/s} \)
2. **Organ Pipe:**
- For an organ pipe with one end closed and the other open, it supports odd harmonics.
- First overtone (second harmonic) frequency formula:
\[
f_1 = \frac{3v}{4L_p} = \frac{3 \times 343}{4 \times 1.7} \approx 151 \text{ Hz}
\]
3. **Violin String:**
- The string's frequency \( f_s \) matches the frequency of the organ pipe's first overtone.
- Fundamental frequency of a string:
\[
f_s = \frac{1}{2L_s}\sqrt{\frac{T}{\mu}}
\]
- Given \( f_s = 151 \text{ Hz} \), we rearrange the formula to solve for tension \( T \):
\[
151 = \frac{1}{2 \times 0.32} \sqrt{\frac{T}{0.0015}}
\]
\[
151 = \frac{1}{0.64} \
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