A fork of frequency 325 Hz is held over a resonance cube appartus and the shortest resounding length of the tube was 0.254 m. What is the next resounding length of the velocity of sound at that temperature was 325 m/s.

Answer

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

### Step 1: Calculate the wavelength (λ)
The wavelength can be calculated using the formula:
\[ \lambda = \frac{c}{f} \]
where:
- \( c \) is the velocity of sound (325 m/s)
- \( f \) is the frequency (325 Hz)

Substituting the values:
\[ \lambda = \frac{325 \, \text{m/s}}{325 \, \text{Hz}} = 1 \, \text{m} \]

### Step 2: Determine the shortest resonating length
The shortest resonating length (first resonating length) is given as 0.254 m.

### Step 3: Calculate the next resonating length
The resonating lengths occur at intervals of \( \frac{\lambda}{2} \). Therefore, we need to find \( \frac{\lambda}{2} \):
\[ \frac{\lambda}{2} = \frac{1 \, \text{m}}{2} = 0.5 \, \text{m} \]

Now, to find the next resonating length, we add \( \frac{\lambda}{2} \) to the first resonating length:
\[ \text{Next resonating length} = \text{First resonating length} + \frac{\lambda}{2} \]
\[ \text{Next resonating length} = 0.254 \, \text{m} + 0.5 \, \text{m} = 0.754 \, \text{m} \]

### Final Answer
The next resonating length is **0.754 m**.

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