One of the harmonic frequencies for a particular string under tension is 310 Hz. The next higher harmonic frequency is 400 Hz. What harmonic frequency is next higher after the hannonc frequency 850 Hz?

Answer

To solve the problem step by step, we will analyze the harmonic frequencies of a string and derive the next higher harmonic frequency after 850 Hz.

### Step 1: Understand the Harmonic Frequencies
The frequencies of the harmonics of a string fixed at both ends can be expressed as:
\[ f_n = \frac{nV}{2L} \]
where:
- \( f_n \) is the frequency of the nth harmonic,
- \( n \) is the harmonic number (1, 2, 3, ...),
- \( V \) is the wave velocity,
- \( L \) is the length of the string.

### Step 2: Set Up the Given Frequencies
From the problem, we know:
- The frequency of the nth harmonic is \( f_n = 310 \, \text{Hz} \).
- The frequency of the (n+1)th harmonic is \( f_{n+1} = 400 \, \text{Hz} \).

### Step 3: Write the Equations for the Given Frequencies
Using the formula for harmonic frequencies:
1. For the nth harmonic:
\[ 310 = \frac{nV}{2L} \quad \text{(Equation 1)} \]
2. For the (n+1)th harmonic:
\[ 400 = \frac{(n+1)V}{2L} \quad \text{(Equation 2)} \]

### Step 4: Find the Difference Between the Two Frequencies
Subtract Equation 1 from Equation 2:
\[ 400 - 310 = \frac{(n+1)V}{2L} - \frac{nV}{2L} \]
This simplifies to:
\[ 90 = \frac{V}{2L} \]
Thus, we find:
\[ \frac{V}{2L} = 90 \, \text{Hz} \quad \text{(Equation 3)} \]

### Step 5: Use the Given Frequency of 850 Hz
Now, we need to find the next harmonic frequency after 850 Hz. Let’s denote the harmonic number corresponding to 850 Hz as \( m \):
\[ 850 = \frac{mV}{2L} \quad \text{(Equation 4)} \]

### Step 6: Substitute the Value of \( \frac{V}{2L} \)
From Equation 3, we substitute \( \frac{V}{2L} = 90 \):
\[ 850 = m \cdot 90 \]
Solving for \( m \):
\[ m = \frac{850}{90} \approx 9.44 \]
Since \( m \) must be an integer, we take \( m = 9 \) (the closest lower integer).

### Step 7: Find the Next Harmonic Frequency
The next harmonic frequency, \( f_{m+1} \), can be calculated using:
\[ f_{m+1} = (m+1) \cdot \frac{V}{2L} \]
Substituting \( m = 9 \) and \( \frac{V}{2L} = 90 \):
\[ f_{m+1} = (9 + 1) \cdot 90 = 10 \cdot 90 = 900 \, \text{Hz} \]

### Step 8: Conclusion
Thus, the next higher harmonic frequency after 850 Hz is:
\[ \boxed{900 \, \text{Hz}} \]