Question

Air is blown at the mouth of a tube of length 25 cm and diameter equal to 2 cm open at both ends. If velocity of sound in air is $330m{s^{ - 1}}$, the sound emitted will have all the frequencies in the group (neglect end correction).
A. $330, 990, 1690 Hz$
B. $302, 664, 1320 Hz$
C. $660,1320,1980 Hz$
D. $660, 1000,3300 Hz$

Answer 1

Hint: The length of the tube is the positive integral multiple of half wavelength of sound waves and velocity is the product of wavelength of sound wave and frequency. We’ll substitute the value of wavelength from the formula of length of the tube in the formula of velocity of the sound wave and then frequency can be calculated. Then, we put the value of n as 1, 2 and 3 in the frequency and all the three frequencies can be calculated.

Complete answer:
Air is blown at the mouth of a tube whose length (l) is 25 cm and diameter (d) is 2 cm and the tube is open from both the ends and thus it is known as open organ tube. As both the ends of the tube are open, therefore the waves are reflected from these ends with change of type but the particles continue to move in the same direction even after the reflection of the waves at the open ends. So, the particles have maximum displacements at the open ends.

Thus, antinodes are formed at the open ends. In other words, antinodes are the position of maximum displacement above and below the rest position. Now, the length of the tube $\left( l \right) = \dfrac{{n\lambda }}{2}$ where λ is the wavelength of the sound and n is the number of nodes formed and also the antinodes formed at the open ends are separated by n nodes and n+1 antinodes. The amplitude whose magnitude is zero is known as node.
Hence,
$l = \dfrac{{n\lambda }}{2}$
$\Rightarrow\lambda = \dfrac{{2l}}{n}$
$\Rightarrow v = \nu \lambda $ where $\nu $ is the frequency and v is the velocity of sound in air.
$\Rightarrow v = \dfrac{{2l\nu }}{n}$
$\Rightarrow\nu = \dfrac{{vn}}{{2l}}$
$\Rightarrow\nu = \dfrac{{n \times 330}}{{2 \times 0.25}}\left[ {1cm = 0.01m} \right]$
$\Rightarrow\nu = 660n$ Hz
Now, when $n = 1$ ,
$\nu = 660 \times 1i.e.,660$ Hz
When $n = 2$ ,
$\nu = 660 \times 2$ i.e., $1320$ Hz
When $n = 3$ ,
$\therefore \nu = 660 \times 3$ i.e., $1980$ Hz
Hence, the sound emitted will have all the frequencies in the group $\left({660,1320,1980Hz} \right)$ .

Therefore, option C is correct.

Note: In an open organ tube or pipe, the tones of frequencies are $\nu ,2\nu ,3\nu ,.......n\nu $ by adjusting the pressure with which air is blown into the tube. The speed of the sound is in $m{s^{ - 1}}$ and hence we should convert the length of the tube from cm to m.