Question

An air column, closed at one end and open at the other end, resonates with a tuning fork of frequency $n$ when its lengths are 45 cm, 99 cm and at two other lengths in between these values. The wavelength of sound in the air column is
(A) 180 cm
(B) 108 cm
(C) 54 cm
(D) 36 cm

Answer 1

Hint : Only odd harmonics are produced in a pipe closed at one end. We can find the initial and the final frequency. Using the value of the length in between 45 cm and 99 cm we can find the wavelength of the air column.

Formula used: The following formulas are used to solve this question.
$L = \dfrac{\lambda }{4}$ , where length is $L$ and wavelength is $\lambda$ .

Complete step by step answer:
A closed-end air column does not possess any even-numbered harmonics. Only odd-numbered harmonics are produced, where the frequency of each harmonic is some odd-numbered multiple of the frequency of the first harmonic. The next highest frequency above the third harmonic is the fifth harmonic.
The distance between adjacent antinodes on a standing wave pattern is equivalent to one-half of a wavelength. Since nodes always lie midway in between the antinodes, the distance between an antinode and a node must be equivalent to one-fourth of a wavelength.
For the first harmonic, the relationship between length and wavelength is given by,
$L = \dfrac{\lambda }{4}$ , where length is $L$ and wavelength is $\lambda$ .
The initial frequency is given by ${f_1} = \dfrac{{nv}}{{4L}}$ where ${f_1}$ is the initial frequency, $L$ is the length, $n$ is the number of harmonics and $v$ is the speed of wave.
The final frequency is given by ${f_2} = \dfrac{{\left( {n + 3} \right)v}}{{4L}}$ where ${f_2}$ is the final frequency, $L$ is the length, $\left( {n + 3} \right)$ is the number of harmonics and $v$ is the speed of wave.
The difference between the two frequencies is given by-
${f_2} - {f_1} = \dfrac{{nv}}{{4L}} - \dfrac{{\left( {n + 3} \right)v}}{{4L}}$
$\Rightarrow {f_2} - {f_1} = \dfrac{{nv - nv + 3v}}{{4L}}$
Thus, ${f_2} - {f_1} = \dfrac{{3v}}{{4L}}$
In the question it is given that, an air column, closed at one end and open at the other end, resonates with a tuning fork of frequency $n$ when its lengths are 45 cm, 99 cm and at two other lengths in between these values.
The difference between these two values is 54cm.
$\therefore$ ${f_2} - {f_1} = \dfrac{{3v}}{{4L}} = 54cm$
Now, we already known that, $\dfrac{v}{{2L}} = \lambda$
Multiplying the same value on both sides,
$\dfrac{3}{2} \times \dfrac{v}{{2L}} = \dfrac{{3v}}{{4L}} = \dfrac{{3\lambda }}{2}$
Thus from the equation, $\dfrac{{3v}}{{4L}} = 54cm$ , we get,
$\dfrac{{3\lambda }}{2} = 54cm$
$\Rightarrow \lambda = 36cm$
The correct answer is option D.

Note:
For an air column, closed at one end and open at the other end, resonates with a tuning fork when its lengths are 45 cm, 99 cm.
Thus we find that both the lengths are multiples of 9, i.e.,
$45 = 5\left( 9 \right)$
and
$99 = 11\left( 9 \right)$
Thus, we can safely conclude that 9 is the fundamental frequency.
Only odd harmonics are produced in a pipe closed at one end. Thus the odd numbers left in between are 7 and 9.
Thus the other lengths are 63cm and 81cm.
We know that, $\dfrac{\lambda }{4} = 9cm$
$\Rightarrow \lambda = 36cm$ .