A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear sound of maximum frequency 15k Hz. What is the highest harmonic that can be played on this string which is audible to the person?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the given values
- Fundamental frequency of the steel wire, \( f_0 = 200 \, \text{Hz} \)
- Maximum frequency audible to the person, \( f' = 15 \, \text{kHz} = 15000 \, \text{Hz} \)

### Step 2: Use the formula for the frequency of harmonics
For a string fixed at both ends, the frequency of the \( n \)-th harmonic is given by:
\[ f_n = n \cdot f_0 \]
where \( f_n \) is the frequency of the \( n \)-th harmonic and \( n \) is the harmonic number.

### Step 3: Set up the equation to find the highest harmonic
We need to find the highest harmonic \( n \) such that:
\[ f_n \leq f' \]
This can be rewritten using the formula for the frequency of harmonics:
\[ n \cdot f_0 \leq f' \]

### Step 4: Solve for \( n \)
Rearranging the equation gives:
\[ n \leq \frac{f'}{f_0} \]
Substituting the values:
\[ n \leq \frac{15000 \, \text{Hz}}{200 \, \text{Hz}} \]

### Step 5: Calculate \( n \)
Now, perform the calculation:
\[ n \leq \frac{15000}{200} = 75 \]

### Step 6: Conclusion
The highest harmonic that can be played on the string which is audible to the person is:
\[ \boxed{75} \]

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