In a resonance tube experiment, when the tube is filled with water up to a height of 17.0 cm from bo

Q: Why is the tube considered a closed organ pipe?

Because the water surface acts as a rigid boundary that reflects the sound waves, creating a node at the water level, just like the closed end of a pipe.

Q: Why is the difference in lengths $\frac{\lambda}{2}$ and not $\frac{\lambda}{4}$?

Consecutive resonances occur at odd multiples of $\frac{\lambda}{4}$ (e.g., $\frac{\lambda}{4}$, $\frac{3\lambda}{4}$, $\frac{5\lambda}{4}$). The difference between any two consecutive odd multiples is $\frac{2\lambda}{4}$, which simplifies to $\frac{\lambda}{2}$.

Q: Does the end correction affect the result?

No, it cancels out! Since the end correction $e$ is added to both resonance lengths, when we take the difference $(l_1 + e) - (l_2 + e)$, the $e$ terms subtract to zero.

Q: Why did the air column length decrease when the water level was raised?

The total length of the physical tube is fixed. When you add more water, the water level rises, leaving less empty space above it for the air column.

Q: What if the speed of sound was not given?

Without the speed of sound, we could only determine the wavelength of the sound wave in the tube. We need the wave speed to calculate the actual frequency using $v = f\lambda$.

Question

In a resonance tube experiment, when the tube is filled with water up to a height of $17.0	ext{ cm}$ from bottom, it resonates with a given tuning fork. When the water level is raised, the next resonance with the same tuning fork occurs at a height of $24.5	ext{ cm}$. If the velocity of sound in air is $330	ext{ m/s}$, the tuning fork frequency is

Options:

(A) $1100	ext{ Hz}$

(B) $3300	ext{ Hz}$

(C) $2200	ext{ Hz}$

(D) $550	ext{ Hz}$

Sigma Percentile · Accepted Answer

Imagine a resonance tube filled with water. The water acts as a rigid boundary, creating a closed organ pipe. The air column above the water vibrates in resonance with a tuning fork. Let's visualize the two resonance states given in the problem.

### The Physics of the Resonance Tube

For a tube closed at one end, resonance happens when the length of the air column fits an odd multiple of quarter wavelengths. A crucial property to remember is that the distance between two consecutive resonances is always exactly half a wavelength.

Notice how raising the water level decreases the air column length. The change in the water level is exactly equal to the change in the air column length. So, the difference between the two water heights, $h_2$ minus $h_1$, must be equal to $\frac{\lambda}{2}$.

### Calculating the Wavelength

Let's substitute the given values into our relationship:

$$\frac{\lambda}{2} = h_2 - h_1$$

$$\frac{\lambda}{2} = 24.5 - 17.0$$

$$\frac{\lambda}{2} = 7.5 	ext{ cm}$$

Multiplying by two, we find that the full wavelength $\lambda$ is $15.0 	ext{ cm}$, or $0.15 	ext{ m}$.

### Finding the Frequency

Now, we need to find the frequency of the tuning fork. We know the universal wave equation:

$$v = f \lambda$$

Rearranging this, frequency $f$ is $v$ divided by $\lambda$. We plug in the speed of sound, $330 	ext{ m/s}$, and our wavelength, $0.15 	ext{ m}$:

$$f = \frac{330}{0.15}$$

Solving this simple division gives us **$2200 	ext{ Hz}$**. This is the natural frequency of our tuning fork.

### What About End Correction?

You might be wondering, what about the end correction $e$? The beauty of the two-resonance method is that when we subtract the two lengths, the end correction perfectly cancels out!

$$(l_1 + e) - (l_2 + e) = l_1 - l_2 = \frac{\lambda}{2}$$

This makes it a highly accurate way to measure the speed of sound or unknown frequencies in the lab without worrying about the exact diameter of the tube.

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Visualized Solution

$Resonance Tube Setup$

$Condition for Resonance$

$Relating Water Levels to Wavelength$

$Calculating Wavelength$

$Calculating Frequency$

$Final Frequency$

$End Correction Consideration$

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Physics of the Resonance Tube

Calculating the Wavelength

Finding the Frequency

What About End Correction?

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resonance Tube Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Resonance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating Water Levels to Wavelength

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Wavelength

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } End Correction Consideration