An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is $50\,cm$. The next larger length of the column resonating with the same tuning fork is :

Question

A steel wire with mass per unit length $ 7.0 \times 10^{-3} \, \text{kg/m} $ is under a tension of $ 70 \, \text{N} $. The speed of transverse waves in the wire will be:

Answer 1

In this problem, we are dealing with the concept of resonance in an air column that is closed at one end and open at the other. Such air columns produce standing waves at specific lengths that correspond to the resonant frequencies of a given tuning fork.

When the air column resonates with a tuning fork, it suggests that the natural frequency of the air column matches the frequency of the tuning fork. In an air column that is closed at one end, the resonant frequencies form a pattern where the length of the air column for resonance is given by:

L_n = \left(n + \frac{1}{2}\right)\frac{\lambda}{4} \, \text{where} \, n = 0, 1, 2, ...

Here, n represents the number of antinodes, and \lambda is the wavelength of the sound wave.

For the smallest length resonating, which is 50 \, \text{cm}, we can say it represents the first resonance (fundamental frequency) or n = 0:

L_0 = \left(0 + \frac{1}{2}\right)\frac{\lambda}{4} = \frac{\lambda}{4}

Given L_0 = 50 \, \text{cm}, we equate and find the wavelength:

\frac{\lambda}{4} = 50 \implies \lambda = 200 \, \text{cm}

The next resonance occurs at n = 1, with length:

L_1 = \left(1 + \frac{1}{2}\right)\frac{\lambda}{4} = \frac{3}{2}\frac{\lambda}{4} = \frac{3\lambda}{4}

Substituting the wavelength found:

L_1 = \frac{3 \times 200}{4} = 150 \, \text{cm}

This confirms that the next larger length of the column resonating with the same tuning fork is 150 cm.

Thus, the correct option is 150 cm, as the pattern of resonance supports longer air columns requiring larger fractions of the wavelength.

Answer 2