Question

A pipe open at both ends of length $1.5\ \text{m}$ is dipped in water such that the second overtone of vibrating air column is resonating with a tuning fork of frequency $330\ \text{Hz}$. If speed of sound in air is $330\ \text{m/s}$, then the length of the pipe immersed in water is (Neglect end correction)

• A. $0.35\ \text{m}$
• B. $0.25\ \text{m}$
• C. $0.55\ \text{m}$
• D. $0.45\ \text{m}$

Hint:

Always carefully note which type of pipe model applies after dipping! Dipping an open pipe into water creates a closed boundary at the fluid surface, completely shifting the frequency system from all integers ($n = 1, 2, 3...$) to odd-only integer harmonics ($n = 1, 3, 5...$). Remembering that the second overtone equals the $5^{\text{th}}$ harmonic for closed pipes keeps your calculations perfectly aligned.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Initially, we have a pipe open at both ends. When it is dipped vertically into water, the water surface seals one end, effectively transforming it into a pipe closed at one end (half-open pipe). The air column remaining above the water resonates in its second overtone with a $330\ \text{Hz}$ tuning fork. We need to find how much of the pipe's length is submerged under water.

Step 2: Key Formula or Approach:
1. For a pipe closed at one end, the resonance frequencies of overtones are odd harmonics of the fundamental frequency. 2. The fundamental frequency is $f_1 = \frac{v}{4L'}$, where $L'$ is the length of the air column above the water. 3. The first overtone is the $3^{\text{rd}}$ harmonic, and the second overtone corresponds to the $5^{\text{th}}$ harmonic: $$f_5 = \frac{5v}{4L'}$$ 4. Once the air column length $L'$ is found, the immersed length $l_{\text{immersed}}$ is determined by subtracting it from the total length $L$: $$l_{\text{immersed}} = L - L'$$

Step 3: Detailed Explanation:
Given values from the problem: Total length of the pipe, $L = 1.5\ \text{m}$ Frequency of the second overtone, $f_5 = 330\ \text{Hz}$ Velocity of sound, $v = 330\ \text{m/s}$ Set up the second overtone frequency formula for a closed-end air column: $$330 = \frac{5 \times 330}{4L'}$$ Divide both sides by $330$: $$1 = \frac{5}{4L'}$$ Isolate $L'$: $$4L' = 5 \implies L' = \frac{5}{4} = 1.25\ \text{m}$$ The length of the active vibrating air column remaining above the water surface is $1.25\ \text{m}$. Now, calculate the length of the pipe that must be immersed under the water: $$l_{\text{immersed}} = L - L' = 1.5\ \text{m} - 1.25\ \text{m} = 0.25\ \text{m}$$

Step 4: Final Answer:
The length of the pipe immersed in water is $0.25\ \text{m}$, which matches option (B).