A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity

Q: Why do we add 'e' to the measured length?

The sound wave doesn't perfectly reflect exactly at the physical boundary of the open tube. The antinode forms slightly outside the tube in the open air. This extra distance is the end correction 'e', making the effective acoustic length longer than the physical length.

Q: Why is the length equal to $\lambda/4$ and not $\lambda/2$?

A resonance tube filled with water acts as a pipe closed at one end (the water surface acts as a rigid boundary, forcing a displacement node) and open at the other (where an antinode forms). The fundamental mode for such a system is a quarter-wavelength, $\lambda/4$.

Q: Could we have solved this by subtracting the two equations instead of dividing?

Yes! You can write $l_1 + e = \frac{v}{4f_1}$ and $l_2 + e = \frac{v}{4f_2}$. Subtracting the first from the second eliminates 'e' immediately: $l_2 - l_1 = \frac{v}{4}(\frac{1}{f_2} - \frac{1}{f_1})$. This is a very powerful and fast alternative method!

Q: Why did we divide by 100 at the very end?

Because our lengths ($11\text{ cm}$ and $27\text{ cm}$) were given in centimeters, the velocity we calculated ($32768$) was in centimeters per second (cm/s). The options are in meters per second (m/s), so we must divide by 100 to convert the units.

Q: Does the jagged end change the fundamental physics of the problem?

Not the physics of the wave itself, but it changes our measurement strategy. Because the end is jagged, we don't know the exact physical length of the tube. The reference mark gives us a consistent starting point, and the unknown 'e' absorbs both the true acoustic end correction and the offset of the reference mark from the 'average' top of the tube.

Question

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency $512	ext{ Hz}$ produces first resonance when the tube is filled with water to a mark $11	ext{ cm}$ below a reference mark near the open end of the tube. The experiment is repeated with another fork of frequency $256	ext{ Hz}$ which produces first resonance when water reaches a mark $27	ext{ cm}$ below the reference mark. The velocity of sound in air, obtained in the experiment is close to

Options:

(A) $328	ext{ ms}^{-1}$

(B) $341	ext{ ms}^{-1}$

(C) $322	ext{ ms}^{-1}$

(D) $335	ext{ ms}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a laboratory with an old resonance tube. The top of the tube is jagged and uneven, which poses a practical problem: we cannot accurately measure the length of the air column from the very top. To overcome this, we use a fixed **reference mark** near the open end.

However, the physics of sound waves tells us that the antinode (the point of maximum displacement) does not form exactly at the physical boundary of the tube. It bulges out slightly into the open air. This extra distance is known as the **end correction**, denoted by $e$. Therefore, the true effective length of our resonating air column is the measured length $l$ (from the reference mark to the water level) plus this unknown end correction $e$.

### The Master Equation

When a tuning fork produces the first resonance in a tube closed at one end (by water), the effective length of the air column $L$ corresponds to one-quarter of the sound's wavelength. Mathematically, this is written as:

$$L = l + e = \frac{\lambda}{4}$$

We also know the fundamental wave equation relating velocity $v$, frequency $f$, and wavelength $\lambda$:

$$v = f\lambda \implies \lambda = \frac{v}{f}$$

Substituting this into our resonance condition gives us the master equation for this experiment:

$$l + e = \frac{v}{4f} \implies v = 4f(l + e)$$

### Setting Up the Two Cases

Let's apply our master equation to the two experimental cases provided. In the first case, we use a tuning fork with a frequency $f_1 = 512	ext{ Hz}$, and the water level is found to be $l_1 = 11	ext{ cm}$ below the reference mark. Plugging these values in, we get:

$$v = 4 	imes 512 	imes (11 + e)$$

$$v = 2048(11 + e) \quad 	ext{... (i)}$$

In the second case, the experiment is repeated with a different tuning fork of frequency $f_2 = 256	ext{ Hz}$. The new resonant length is $l_2 = 27	ext{ cm}$. Substituting these new values yields:

$$v = 4 	imes 256 	imes (27 + e)$$

$$v = 1024(27 + e) \quad 	ext{... (ii)}$$

### Solving for the End Correction

Here is the crucial insight: the velocity of sound $v$ in the air inside the laboratory remains constant across both experiments. This allows us to safely equate equation (i) and equation (ii):

$$2048(11 + e) = 1024(27 + e)$$

Notice how elegantly the numbers are chosen! $2048$ is exactly double $1024$. Dividing both sides by $1024$ drastically simplifies our algebra:

$$2(11 + e) = 27 + e$$

Expanding the left side and rearranging the terms to isolate $e$:

$$22 + 2e = 27 + e$$

$$2e - e = 27 - 22$$

$$e = 5	ext{ cm}$$

We have successfully found the end correction!

### Final Calculation

With the end correction $e = 5	ext{ cm}$ in hand, we can substitute it back into either of our initial velocity equations. Let's use equation (i):

$$v = 2048(11 + 5)$$

$$v = 2048 	imes 16$$

$$v = 32768	ext{ cm/s}$$

Since the options are given in standard SI units (meters per second), we must convert our result by dividing by $100$:

$$v = \frac{32768}{100} = 327.68	ext{ m/s}$$

Rounding off to the nearest integer, we get our final answer:

**$v \approx 328	ext{ m/s}$**

Select Answer:

Visualized Solution

Analyzing the Setup

First Resonance Condition

Equation for Case 1

Equation for Case 2

Equating Velocities

Solving for End Correction ( $e$ )

Calculating Velocity

Final Conversion

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

Setting Up the Two Cases

Solving for the End Correction

Final Calculation

Similar Questions

A tuning fork of frequency $480 Hz$ is used in an experiment for measuring speed of sound ( $v$ ) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column $l_{1} = 30 cm$ and $l_{2} = 70 cm$ . Then, $v$ is equal to

A tuning fork of $512 Hz$ is used to produce resonance in a resonance tube experiment. The level of water at first resonance is $30.7 cm$ and at second resonance is $63.2 cm$ . The error in calculating velocity of sound is (assume actual speed of sound in air is $330 m/s$ ):

An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency $264 Hz$ , if the length of the column in cm is (Speed of sound in air = $330 m/s$ )

Select Answer:

Visualized Solution

Analyzing the Setup

First Resonance Condition

Equation for Case 1

Equation for Case 2

Equating Velocities

Solving for End Correction ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e)

Calculating Velocity

Final Conversion

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

Setting Up the Two Cases

Solving for the End Correction

Final Calculation

Similar Questions

Solving for End Correction ( $e$ )