To solve this problem, we will calculate the wave speed on the piano string, the length of the string for the fundamental frequency, and the length of the closed-open pipe that produces the same fundamental frequency.

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Part 1: Wave Speed on the Piano String

The wave speed $v$ on a string is determined using the formula:

$v = \sqrt{\frac{T}{\mu}}$

Where:

• $T = 750 \, \text{N}$ (tension),
• $\mu = 3.4 \times 10^{-3} \, \text{kg/m}$ (mass per unit length).

$v = \sqrt{\frac{750}{3.4 \times 10^{-3}}}$

$v = \sqrt{220588.24} \approx 469.62 \, \text{m/s}$

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Part 2: Length of the String for Fundamental Frequency

The fundamental frequency $f$ for a string is given by:

$f = \frac{v}{2L}$

Rearranging to find the length $L$:

$L = \frac{v}{2f}$

Substitute:

• $v = 469.62 \, \text{m/s}$,
• $f = 262 \, \text{Hz}$:

$L = \frac{469.62}{2 \times 262}$

$L \approx 0.896 \, \text{m}$

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Part 3: Length of the Closed-Open Tube

The fundamental frequency $f$ for a pipe closed at one end is given by:

$f = \frac{v_{\text{air}}}{4L_{\text{pipe}}}$

Where $v_{\text{air}}$ is the speed of sound in air. Assuming standard conditions:

$v_{\text{air}} = 343 \, \text{m/s}$

Rearranging to find $L_{\text{pipe}}$:

$L_{\text{pipe}} = \frac{v_{\text{air}}}{4f}$

Substitute:

• $v_{\text{air}} = 343 \, \text{m/s}$,
• $f = 262 \, \text{Hz}$:

$L_{\text{pipe}} = \frac{343}{4 \times 262}$

$L_{\text{pipe}} \approx 0.327 \, \text{m}$

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Summary of Results

1. Wave Speed on the String: $v \approx 469.62 \, \text{m/s}$.
2. Length of the String: $L \approx 0.896 \, \text{m}$.
3. Length of the Closed-Open Tube: $L_{\text{pipe}} \approx 0.327 \, \text{m}$.

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Would you like further clarifications or a deeper dive into these concepts?

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Relative Questions

1. What is the relationship between tension and wave speed on a string?
2. How does the length of a closed-open tube affect its fundamental frequency?
3. Why does the wave speed depend on the mass per unit length in a string?
4. What are some real-world factors that might affect the speed of sound in air?
5. How would the results change if the pipe were open at both ends?

Tip

For a string, increasing tension increases wave speed, while increasing mass per