A 60.0 cm guitar string under a tension of 50.0 N has a mass per unit length of 0.100 `g//cm.` What is the highest resonance frequency of the string that can be heard by a person able to hear frequencies upto 20000 Hz?

Explanation

To solve the problem step by step, we need to follow these calculations:

### Step 1: Convert mass per unit length to SI units
The mass per unit length (μ) is given as 0.100 g/cm. We need to convert this to kg/m:
- 0.100 g/cm = 0.100 g/cm × (1 kg/1000 g) × (100 cm/1 m) = 0.100 × 10^-3 kg/10^-2 m = 0.001 kg/m

### Step 2: Calculate the wave speed (V) on the string
The wave speed (V) on a string under tension is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
Where:
- T = tension in the string = 50 N
- μ = mass per unit length = 0.001 kg/m

Substituting the values:
\[ V = \sqrt{\frac{50 \, \text{N}}{0.001 \, \text{kg/m}}} = \sqrt{50000} \approx 223.61 \, \text{m/s} \]

### Step 3: Calculate the fundamental frequency (f1)
The fundamental frequency (f1) of a string fixed at both ends is given by:
\[ f_1 = \frac{V}{2L} \]
Where:
- L = length of the string = 60 cm = 0.6 m

Substituting the values:
\[ f_1 = \frac{223.61 \, \text{m/s}}{2 \times 0.6 \, \text{m}} = \frac{223.61}{1.2} \approx 186.34 \, \text{Hz} \]

### Step 4: Determine the highest harmonic frequency (fn)
The frequency of the nth harmonic is given by:
\[ f_n = n \cdot f_1 \]
We need to find the maximum n such that \( f_n \leq 20000 \, \text{Hz} \):
\[ n \cdot 186.34 \leq 20000 \]
\[ n \leq \frac{20000}{186.34} \approx 107.3 \]

Since n must be an integer, the maximum n is 107.

### Step 5: Calculate the highest resonance frequency (f_max)
Now, we can calculate the highest resonance frequency:
\[ f_{\text{max}} = n \cdot f_1 = 107 \cdot 186.34 \approx 19977.58 \, \text{Hz} \]

### Conclusion
The highest resonance frequency of the string that can be heard by a person is approximately **19977.58 Hz**.

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