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:

We are dealing with a resonance tube that is closed at the bottom and open at the top. Such an air column behaves like a closed organ pipe. For a closed organ pipe, the resonant (standing-wave) lengths of the air column are given by the well-known quarter-wavelength formula

$$l_n \;=\;\frac{(2n-1)\,\lambda}{4},\qquad n = 1,2,3,\ldots$$

where $$l_n$$ is the length of the air column at the $$n^{\text{th}}$$ resonance and $$\lambda$$ is the wavelength of sound in air.

Successive resonances correspond to consecutive odd multiples of $$\lambda/4$$. Therefore, the difference between any two successive resonant lengths is

$$l_{n+1}-l_n=\frac{(2(n+1)-1)\lambda}{4}-\frac{(2n-1)\lambda}{4} =\frac{(2n+1)\lambda}{4}-\frac{(2n-1)\lambda}{4} =\frac{\lambda}{2}.$$

This result is very useful because the end correction (the small extra length added because the antinode forms slightly above the tube’s rim) cancels out when we take the difference. Hence, from two successive resonances we can get $$\lambda$$ directly.

In the experiment we are told that the first observed resonant length is

$$l_1 = 30\ \text{cm} = 0.30\ \text{m},$$

and the next (successive) resonant length is

$$l_2 = 70\ \text{cm} = 0.70\ \text{m}.$$

We now calculate the difference:

$$l_2 - l_1 = 0.70\ \text{m} - 0.30\ \text{m} = 0.40\ \text{m}.$$

But we have just established that

$$l_2 - l_1 = \frac{\lambda}{2}.$$

So we can write

$$\frac{\lambda}{2} = 0.40\ \text{m}.$$

Multiplying both sides by 2 gives the full wavelength:

$$\lambda = 2 \times 0.40\ \text{m} = 0.80\ \text{m}.$$

Now we use the basic wave relation between speed $$v$$, frequency $$f$$ and wavelength $$\lambda$$. The formula is

$$v = f \lambda.$$

The frequency of the tuning fork has been provided as

$$f = 480\ \text{Hz}.$$

Substituting $$f = 480\ \text{Hz}$$ and $$\lambda = 0.80\ \text{m}$$ into the wave equation, we get

$$v = (480\ \text{Hz})(0.80\ \text{m}) = 384\ \text{m s}^{-1}.$$

Hence, the correct answer is Option A.