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:
A vibrating tuning fork with a fixed frequency $n$ drives an air column inside a tube closed at one end. We are looking for the velocity of sound in terms of the two consecutive resonant air column lengths, $L_1$ (first harmonic) and $L_2$ (third harmonic).

Step 2: Key Formula or Approach:
For an air column closed at one end, resonance occurs at odd quarter-wavelength multiples: $$\text{First resonance: } L_1 = \frac{\lambda}{4}$$ $$\text{Second consecutive resonance: } L_2 = \frac{3\lambda}{4}$$ By tracking the difference between these two consecutive boundary modes, we eliminate potential end-correction errors and find $\lambda$. Then, we substitute $\lambda$ into the wave speed formula: $$v = n\lambda$$

Step 3: Detailed Explanation:
Let's find the difference between our two consecutive resonant length constraints: $$L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4}$$ $$L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2}$$ Isolate the wavelength variable $\lambda$: $$\lambda = 2(L_2 - L_1)$$ Now, substitute this value into our fundamental velocity of sound equation: $$v = n \cdot [2(L_2 - L_1)] = 2n(L_2 - L_1)$$

Step 4: Final Answer:
The speed of sound in air is $2n(L_2 - L_1)$, which matches option (C).