A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of wa

Q: Why is the first resonance water level higher than the second resonance water level?

The first resonance (fundamental mode) requires a shorter air column ($̵/4$) compared to the second resonance ($3̵/4$). Since the air column length is $L - y$, a shorter air column directly corresponds to a higher water level $y$ from the bottom.

Q: Does the end correction affect the calculation of the wavelength in this problem?

No. When we subtract the two resonance equations, the end correction $e$ (along with the total tube length $L$) cancels out completely, leaving only the difference in water levels equal to half the wavelength.

Q: What is the physical meaning of end correction?

End correction accounts for the fact that the pressure variations do not drop to zero exactly at the open end of the tube, but slightly outside it (typically around $0.6r$, where $r$ is the tube radius).

Q: Why did we assume the actual speed of sound is 330 m/s?

The actual speed of sound of $330\text{ m/s}$ is a standard value for air at room temperature and is explicitly provided in standard versions of this JEE problem to calculate the error.

Q: Can we find the total length of the tube $L$ from the given data?

No, we cannot find the total length $L$ unless either the end correction $e$ or the tube radius $r$ is explicitly provided.

Question

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

Options:

(A) 204.1	ext{ cm/s}

(B) 110	ext{ cm/s}

(C) 58	ext{ cm/s}

(D) 280	ext{ cm/s}

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In a standard **resonance column experiment**, a tuning fork of known frequency $f$ is held over the open end of a vertical tube containing water. The water surface acts as a closed, rigid boundary, forcing a **displacement node** to form at the water-air interface. The open top of the tube acts as a **displacement antinode**.

Let $L$ be the total length of the tube, and let the water level measured from the bottom of the tube be $y$. The length of the vibrating air column is therefore:

$$l = L - y$$

### The Water Level Trap

Many students fall into a classic trap by assuming that the given values $30.7	ext{ cm}$ and $63.2	ext{ cm}$ represent the lengths of the air column ($l_1$ and $l_2$). However, the problem explicitly states these are the **levels of water** ($y_1$ and $y_2$) measured from the bottom.

Let's analyze the relationship between the air column length and the water level:
- The **first resonance** (fundamental mode) occurs at the shortest possible air column length:
  $$l_1 + e = \frac{\lambda}{4}$$
- The **second resonance** (first overtone) occurs at the next resonance length:
  $$l_2 + e = \frac{3\lambda}{4}$$

Since $l_1 < l_2$, the corresponding water level for the first resonance ($y_1$) must be **higher** than the water level for the second resonance ($y_2$):

$$y_1 = 63.2	ext{ cm} \quad 	ext{and} \quad y_2 = 30.7	ext{ cm}$$

### Formulating the Equations

Including the end correction $e$ at the open end, we can write the effective lengths for both resonances:

$$L - y_1 + e = \frac{\lambda}{4} \quad 	ext{--- (Equation 1)}$$

$$L - y_2 + e = \frac{3\lambda}{4} \quad 	ext{--- (Equation 2)}$$

To eliminate the unknown tube length $L$ and the end correction $e$, we subtract Equation 1 from Equation 2:

$$(L - y_2 + e) - (L - y_1 + e) = \frac{3\lambda}{4} - \frac{\lambda}{4}$$

$$y_1 - y_2 = \frac{\lambda}{2}$$

### Calculating Wavelength and Velocity

Substituting the given water levels into our simplified equation:

$$63.2	ext{ cm} - 30.7	ext{ cm} = \frac{\lambda}{2}$$

$$32.5	ext{ cm} = \frac{\lambda}{2} \implies \lambda = 65.0	ext{ cm} = 0.65	ext{ m}$$

Now, we calculate the observed speed of sound ($v_{	ext{obs}}$) using the wave speed formula:

$$v_{	ext{obs}} = f \lambda = 512	ext{ Hz} 	imes 0.65	ext{ m} = 332.8	ext{ m/s}$$

### Finding the Error

The actual speed of sound in air is given as $330	ext{ m/s}$. The error in our calculated velocity is:

$$\Delta v = v_{	ext{obs}} - v_{	ext{actual}} = 332.8	ext{ m/s} - 330	ext{ m/s} = 2.8	ext{ m/s}$$

Converting this error to centimeters per second:

$$\Delta v = 2.8 	imes 100	ext{ cm/s} = \mathbf{280	ext{ cm/s}}$$

Thus, the correct option is **(d)**.

Select Answer:

Visualized Solution

Visualizing the Resonance Tube Setup

Identifying the Water Level Trap

Setting up the Resonance Equations

Eliminating Unknowns by Subtraction

Calculating the Wavelength $λ$

Calculating the Observed Speed of Sound

Determining the Error in Velocity

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Water Level Trap

Formulating the Equations

Calculating Wavelength and Velocity

Finding the Error

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

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 = $330 m/s$ )