Question

In fundamental mode, the time required for the sound wave to reach up to closed end of a pipe filled with air is '\(t\)' second. The frequency of vibration of air column is (Neglect end correction)

• A. \((4t)^{-1}\)
• B. \((2t)^{-1}\)
• C. \(4t\)
• D. \(2t\)

Hint:

Time for one full wave is the period $T$. In a closed pipe, the wave travels $L$ (which is $1/4$ of $\lambda$), so $t = T/4$. Thus $n = 1/T = 1/4t$.

Answer 1

The Correct Option is A

Step 1: Concept
In a closed pipe, the length $L$ is related to wavelength in fundamental mode by $L = \lambda/4$.

Step 2: Meaning
The time taken to travel distance $L$ with velocity $v$ is $t = L/v$.

Step 3: Analysis
Frequency $n = v/\lambda$.
Substitute $\lambda = 4L$: $n = v/4L$.
From $t = L/v$, we get $1/t = v/L$.
So, $n = \frac{1}{4} \times \frac{v}{L} = \frac{1}{4t} = (4t)^{-1}$.

Step 4: Conclusion
The frequency is $(4t)^{-1}$. Final Answer: (A)