While measuring the speed of sound by performing a resonance column experiment, a student gets the f

Q: Why is the second resonance frequency $3v/4L$?

In a closed pipe, only odd harmonics are present. The first resonance is the fundamental ($v/4L$), and the second resonance is the first overtone, which corresponds to the 3rd harmonic ($3v/4L$).

Q: Why do we equate the frequencies?

The frequency of the sound wave is determined entirely by the source, which is the tuning fork. Since the same tuning fork is used in both seasons, the frequency remains constant regardless of the medium's temperature.

Q: How does temperature affect the speed of sound?

The speed of sound in an ideal gas is given by $v = \sqrt{\gamma RT / M}$. Therefore, $v \propto \sqrt{T}$, meaning sound travels faster in warmer air.

Q: What if the experiment was done in the exact same season?

If the temperature was constant, $v_1$ would equal $v_2$, and the second resonance length $x$ would be exactly $3 \times 18 = 54 \text{ cm}$.

Q: Does the density of air change with temperature?

Yes, as temperature increases, air expands and its density decreases. This decrease in density is the physical reason why the speed of sound increases in warmer air.

Question

While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of $18	ext{ cm}$ during winter. Repeating the same experiment during summer, she measures the column length to be $x	ext{ cm}$ for the second resonance. Then,

Options:

(A) $18 > x$

(B) $x > 54$

(C) $54 > x > 36$

(D) $36 > x > 18$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Seasons
Imagine you are standing in a physics laboratory, conducting the classic resonance column experiment. The goal is to measure the speed of sound in air.

The apparatus consists of a long cylindrical tube partially filled with water. The water surface acts as a rigid closed end, while the top of the tube is open to the atmosphere. By striking a tuning fork and holding it over the open end, we can create standing sound waves inside the tube.

The problem presents us with two distinct scenarios: a cold winter day and a hot summer day. We are tasked with finding the relationship between the resonance lengths in these two seasons.

### The Winter Measurement
During the winter, the student finds the **first resonance** at a column length of $l_1 = 18	ext{ cm}$.

The first resonance corresponds to the fundamental mode of vibration in a closed organ pipe. In this mode, there is a single node at the water surface and a single antinode at the open end. The length of the air column is exactly one-quarter of the wavelength, $l_1 = \frac{\lambda}{4}$.

Since the wave equation relates speed, frequency, and wavelength as $v = f\lambda$, we can express the frequency of the tuning fork as:
$$f = \frac{v_1}{4l_1}$$
Here, $v_1$ represents the speed of sound in the cold winter air.

### The Summer Measurement
Fast forward to the summer. The student repeats the experiment using the **exact same tuning fork**. This time, she is looking for the **second resonance**, which she finds at a new length, $l_2 = x	ext{ cm}$.

The second resonance in a closed pipe corresponds to the first overtone, or the third harmonic. The wave pattern now contains two nodes and two antinodes, and the length of the air column is three-quarters of the wavelength, $l_2 = \frac{3\lambda}{4}$.

The frequency for this mode is given by:
$$f = \frac{3v_2}{4l_2}$$
Here, $v_2$ represents the speed of sound in the warm summer air.

### The Master Equation
Here is the crucial logical bridge: the frequency of the sound is determined entirely by the source, which is the tuning fork. Because the student used the same tuning fork in both seasons, the frequency $f$ remains absolutely constant.

We can confidently equate our two frequency expressions:
$$\frac{v_1}{4l_1} = \frac{3v_2}{4l_2}$$

Our goal is to find the summer resonance length, $l_2$ (which is $x$). Let's rearrange the equation to isolate $l_2$:
$$l_2 = 3l_1 \left( \frac{v_2}{v_1} ight)$$

Substituting the known winter length $l_1 = 18	ext{ cm}$, we get:
$$x = 3(18) \left( \frac{v_2}{v_1} ight)$$
$$x = 54 \left( \frac{v_2}{v_1} ight)$$

### The Hidden Variable: Temperature
If the speed of sound were constant, $x$ would be exactly $54	ext{ cm}$. But we must remember our thermodynamics!

The speed of sound in an ideal gas is directly proportional to the square root of its absolute temperature:
$$v \propto \sqrt{T}$$

Since the summer temperature is strictly greater than the winter temperature ($T_{	ext{summer}} > T_{	ext{winter}}$), the speed of sound in summer must be greater than the speed of sound in winter:
$$v_2 > v_1$$

This implies that the ratio of the speeds is greater than one:
$$\frac{v_2}{v_1} > 1$$

### The Final Deduction
We are at the finish line. We have our equation for $x$, and we know the behavior of the velocity ratio.

Since we are multiplying $54$ by a number strictly greater than $1$, the result must be greater than $54$:
$$x = 54 \left( \frac{v_2}{v_1} ight) > 54(1)$$
$$x > 54$$

The new resonance length $x$ must be greater than $54	ext{ cm}$. This problem is a beautiful demonstration of how mechanical waves and thermodynamics intertwine in the real world!

Select Answer:

Visualized Solution

The Resonance Column

Winter: First Resonance

Summer: Second Resonance

Equating Frequencies

Solving for $x$

The Temperature Effect

Final Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Setup

The Winter Measurement

The Summer Measurement Fast forward to the summer

The Master Equation Here is the crucial logical bridge: the frequency of the sound is determined entirely by the source, which is the tuning fork

The Hidden Variable

The Final Deduction We are at the finish line

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