A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain heigh

Q: Why does the air column behave like a closed organ pipe if the tube is open at both ends?

Even though the physical tube is open at both ends, the presence of water inside seals the bottom end of the air column. Because air molecules cannot vibrate freely at the water-air interface, this boundary acts as a displacement node, making the air column behave exactly like a closed organ pipe.

Q: Why do we neglect end correction in this problem?

The problem explicitly states to 'neglect end correction.' In practice, end correction adds a small effective length (approximately $0.6r$) to the open end of the pipe, but ignoring it simplifies the calculations without changing the core physics being tested.

Q: How do we know which water levels correspond to the 'first' and 'second' resonances?

The pipe is initially filled to the top ($H \approx 3.6\text{ m}$), meaning the air column is initially very short. As water leaks out, the water level $H$ decreases, and the air column length $l$ increases. The first resonance occurs at the highest water level ($H = 3.2\text{ m}$), and the second occurs as the water level drops to the next resonant height ($H = 2.4\text{ m}$).

Q: Why is the rate of fall of the water level not constant?

According to Torricelli's Law, the velocity of water exiting the hole depends on the square root of the water height ($H$). As the water level drops, the pressure at the bottom decreases, which slows down the efflux velocity and consequently reduces the rate at which the water level falls.

Q: What would happen to the resonance heights if the temperature of the air increased?

An increase in air temperature increases the speed of sound ($v \propto \sqrt{T}$). Since the resonant wavelengths are given by $\lambda = v/f$, a higher speed of sound would result in longer resonant air columns, meaning the water levels at which resonance occurs would be lower.

Question

A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain heights in the pipe. Find the heights of water level (from the bottom of the pipe) at which resonances occur. Neglect end correction. Now the pipe is filled to a height $H (\approx 3.6	ext{ m})$. A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of $H$. If the radii of the pipe and the hole are $2 	imes 10^{-2}	ext{ m}$ and $1 	imes 10^{-3}	ext{ m}$ respectively. Calculate the time interval between the occurrence of first two resonances. Speed of sound in air is $340	ext{ m/s}$ and $g = 10	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing next to a tall, vertical cylindrical tube filled with water.

At the top, the tube is open to the air, while the bottom is sealed by the water surface.

This system forms a classic **closed organ pipe** of variable length.

The length of the air column, which we will denote as $l$, is determined by the total length of the pipe $L = 3.6	ext{ m}$ and the height of the water level $H$:

$$l = L - H = 3.6 - H$$

When a tuning fork of frequency $f = 212.5	ext{ Hz}$ is held over the open end, it sends sound waves down the tube.

These waves reflect off the water surface, creating a standing wave pattern.

Resonance occurs when the length of the air column matches one of the natural modes of vibration of the closed pipe.

### The Resonance Condition

For a pipe closed at one end, the boundary conditions require a **displacement node** at the water surface (where air molecules cannot move) and a **displacement antinode** at the open end (where air molecules vibrate with maximum amplitude).

This constraint means that the length of the air column must be an odd multiple of a quarter-wavelength:

$$l_n = (2n-1)\frac{\lambda}{4} = (2n-1)\frac{v}{4f}$$

where $n = 1, 2, 3, \dots$ represents the harmonic mode, and $v = 340	ext{ m/s}$ is the speed of sound in air.

Let us first find the fundamental resonant length $l_0$ (for $n=1$):

$$l_0 = \frac{v}{4f} = \frac{340}{4 	imes 212.5} = 0.4	ext{ m}$$

Using this fundamental length, the possible resonant air column lengths are:

$$l_n = (2n-1) 	imes 0.4	ext{ m}$$

Since the physical length of the pipe is $3.6	ext{ m}$, the air column length $l_n$ cannot exceed $3.6	ext{ m}$.

This gives us the following possible resonant air column lengths:

$$l_n \in \{0.4	ext{ m}, 1.2	ext{ m}, 2.0	ext{ m}, 2.8	ext{ m}, 3.6	ext{ m}\}$$

To find the corresponding water heights $H_n$ from the bottom of the pipe, we subtract each $l_n$ from the total length $L = 3.6	ext{ m}$:

$$H_n = 3.6 - l_n$$

This yields the water heights at which resonance occurs:

$$\mathbf{H \in \{3.2	ext{ m}, 2.4	ext{ m}, 1.6	ext{ m}, 0.8	ext{ m}, 0	ext{ m}\}}$$

### Fluid Dynamics of the Leaking Pipe

Now, let us introduce a dynamic twist to the problem.

The pipe is filled to the top ($H \approx 3.6	ext{ m}$), and a small hole is drilled at the very bottom.

Water begins to leak out, causing the water level to fall.

To describe how fast the water level drops, we must combine two fundamental principles of fluid mechanics: **Torricelli's Law** and the **Equation of Continuity**.

According to Torricelli's Law, the velocity of efflux $v_{	ext{efflux}}$ of water leaving a hole under a head of height $H$ is:

$$v_{	ext{efflux}} = \sqrt{2gH}$$

By the Equation of Continuity, the volume flow rate of water leaving the top of the pipe must equal the volume flow rate leaving the hole at the bottom:

$$A \left(-\frac{dH}{dt}ight) = a v_{	ext{efflux}}$$

where $A = \pi R^2$ is the cross-sectional area of the pipe, and $a = \pi r^2$ is the cross-sectional area of the hole.

Rearranging this equation gives the rate of fall of the water level:

$$-\frac{dH}{dt} = \frac{a}{A} \sqrt{2gH} = \left(\frac{r}{R}ight)^2 \sqrt{2gH}$$

Substituting the given values ($r = 1 	imes 10^{-3}	ext{ m}$, $R = 2 	imes 10^{-2}	ext{ m}$, and $g = 10	ext{ m/s}^2$):

$$\left(\frac{r}{R}ight)^2 = \left(\frac{10^{-3}}{2 	imes 10^{-2}}ight)^2 = \left(\frac{1}{20}ight)^2 = \frac{1}{400}$$

$$-\frac{dH}{dt} = \frac{1}{400} \sqrt{2 	imes 10 	imes H} = \frac{\sqrt{20}}{400} \sqrt{H}$$

Since $\sqrt{20} \approx 4.472$:

$$\mathbf{-\frac{dH}{dt} \approx (1.11 	imes 10^{-2}) \sqrt{H}}$$

This elegant differential equation describes the rate of fall of the water level as a function of its instantaneous height $H$.

### Calculating the Time Interval

As the water level falls from $3.6	ext{ m}$, the first resonance is heard when the water level reaches $H_1 = 3.2	ext{ m}$.

The second resonance is heard when the water level falls further to $H_2 = 2.4	ext{ m}$.

To find the time interval $\Delta t$ between these two events, we must integrate our differential equation between these limits:

$$H^{-1/2} dH = -1.118 	imes 10^{-2} dt$$

Integrating both sides:

$$\int_{3.2}^{2.4} H^{-1/2} dH = -1.118 	imes 10^{-2} \int_0^t dt$$

$$\left[2\sqrt{H}ight]_{3.2}^{2.4} = -1.118 	imes 10^{-2} t$$

$$2(\sqrt{3.2} - \sqrt{2.4}) = 1.118 	imes 10^{-2} t$$

Let us calculate the numerical values of the square roots:

$$\sqrt{3.2} \approx 1.789	ext{ m}^{1/2}$$

$$\sqrt{2.4} \approx 1.549	ext{ m}^{1/2}$$

Substituting these back into the equation:

$$2(1.789 - 1.549) = 1.118 	imes 10^{-2} t$$

$$2(0.240) = 1.118 	imes 10^{-2} t$$

$$0.480 = 1.118 	imes 10^{-2} t$$

Solving for $t$:

$$t = \frac{0.480}{1.118 	imes 10^{-2}} \approx 42.93	ext{ s}$$

Rounding to the nearest integer, we find that the time interval between the first two resonances is **$43	ext{ s}$**.

Visualized Solution

Visualizing the Resonance Column

Resonance Condition for Closed Pipe

Calculating the Fundamental Length $l_{0}$

Determining Resonant Air Column Lengths

Calculating Water Levels

Torricelli's Law & Continuity

Deriving the Rate of Fall $- \frac{d H}{d t}$

Simplifying the Rate of Fall

Setting up Integration for Time Interval

Solving the Definite Integral

Numerical Calculation

Final Value of Time Interval

Summary of Final Answers

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Resonance Condition

Fluid Dynamics of the Leaking Pipe

Calculating the Time Interval

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