Question

A tuning fork of frequency '$n$' is held near the open end of a tube which is closed at the other end and the lengths are adjusted until resonance occurs. The first resonance occurs at length $L_1$ and the immediate next resonance occurs at length $L_2$. The speed of sound in air is

• A. $n(L_2 - L_1)$
• B. $\frac{n(L_2 - L_1)}{2}$
• C. $2n(L_2 - L_1)$
• D. $\frac{n(L_2 + L_1)}{2}$

Hint:

The distance between any two consecutive nodes (or consecutive resonance states) in a standing wave column is always exactly equal to half a wavelength ($\frac{\lambda}{2}$). Therefore, you can directly write $L_2 - L_1 = \frac{\lambda}{2} \implies \lambda = 2\Delta L$, bypassing individual mode equations entirely!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the speed of sound v in terms of tuning fork frequency n and two consecutive resonant lengths L₁ and L₂ for a closed pipe.

Step 2: Key Formula or Approach:
For a closed pipe, L₁ = λ/4, L₂ = 3λ/4. Subtracting eliminates end correction: L₂–L₁ = λ/2. Then v = nλ.

Step 3: Detailed Explanation:
L₂–L₁ = λ/2 → λ = 2(L₂–L₁). Substituting into v = nλ gives v = 2n(L₂–L₁).

Step 4: Final Answer:
Speed of sound is 2n(L₂ – L₁), matching option (C).