A string fastened at both ends has successive resonances with wavelengths of 0.54 m for nth harmonic and 0.48 m for the (n+1) th harmonic and 0.48 m for the (n+1) th harmonic.
(a) Which harmonics are these?
(b) What is the length of the stirng?
(c) What is the wavelength of the fundamental frequency?

Explanation

Let's solve the problem step by step.

### Step 1: Understanding the relationship between wavelength and harmonics
For a string fixed at both ends, the relationship between the length of the string (L), the wavelength (λ), and the harmonic number (n) is given by the formula:
\[ L = \frac{n \lambda}{2} \]
This implies that the wavelength is inversely proportional to the harmonic number:
\[ \lambda_n = \frac{2L}{n} \]

### Step 2: Setting up the equations for the nth and (n+1)th harmonics
Given the wavelengths for the nth and (n+1)th harmonics:
- For nth harmonic: \( \lambda_n = 0.54 \, \text{m} \)
- For (n+1)th harmonic: \( \lambda_{n+1} = 0.48 \, \text{m} \)

Using the relationship:
\[ \frac{n+1}{n} = \frac{\lambda_n}{\lambda_{n+1}} \]
Substituting the given values:
\[ \frac{n+1}{n} = \frac{0.54}{0.48} \]

### Step 3: Solving for n
Cross-multiplying gives:
\[ 0.54n = 0.48(n + 1) \]
Expanding the right-hand side:
\[ 0.54n = 0.48n + 0.48 \]
Rearranging the equation:
\[ 0.54n - 0.48n = 0.48 \]
\[ 0.06n = 0.48 \]
Dividing both sides by 0.06:
\[ n = \frac{0.48}{0.06} = 8 \]

### Step 4: Finding the length of the string
Now that we have \( n = 8 \), we can find the length of the string using the wavelength of the nth harmonic:
\[ L = \frac{n \lambda_n}{2} \]
Substituting the values:
\[ L = \frac{8 \times 0.54}{2} \]
Calculating:
\[ L = \frac{4.32}{2} = 2.16 \, \text{m} \]

### Step 5: Finding the wavelength of the fundamental frequency
The fundamental frequency corresponds to the first harmonic (n=1):
\[ \lambda_1 = \frac{2L}{1} = 2L \]
Substituting the length we found:
\[ \lambda_1 = 2 \times 2.16 = 4.32 \, \text{m} \]

### Final Answers
(a) The harmonics are \( n = 8 \) and \( n + 1 = 9 \).
(b) The length of the string is \( 2.16 \, \text{m} \).
(c) The wavelength of the fundamental frequency is \( 4.32 \, \text{m} \).

---