A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of $27^\circ\text{C}$ two successive resonances are produced at $20 \text{ cm}$ and $73 \text{ cm}$ of column length. If the frequency of the tuning fork is $320 \text{ Hz}$, the velocity of sound in air at $27^\circ\text{C}$ is:
The two nearest harmonics of a tube closed at one end and open at other end are $220 \text{ Hz}$ and $260 \text{ Hz}$. What is the fundamental frequency of the system?
(2017-Delhi)
For a closed pipe, frequencies are odd multiples of fundamental $f_0$. Difference between successive harmonics is $2f_0 = 260 - 220 = 40 \text{ Hz}$. Thus, $f_0 = 20 \text{ Hz}$.
The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ meter long. The length of the open pipe will be:
(2016-II)
2nd overtone of open pipe = $3 \times (v / 2L_{\text{open}})$. 1st overtone of closed pipe = $3 \times (v / 4L)$. Equating them: $3v / 2L_{\text{open}} = 3v / 4L \Rightarrow L_{\text{open}} = 2L$.
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 $50 \text{ cm}$. The next larger length of the column resonating with the same tuning fork is:
If we study the vibration of a pipe open at both ends, then the following statement is not true:
(2013)
In an open organ pipe, the open ends act as displacement antinodes and pressure nodes. Therefore, the pressure change is minimum (zero) at both ends, making statement 'a' incorrect.
If the tension and diameter of a sonometer wire of fundamental frequency $n$ is doubled and density is halved then its fundamental frequency will become:
(2001)
Fundamental frequency $n = \frac{1}{lD} \sqrt{\frac{T}{\pi \rho}}$. With new values $T' = 2T$, $D' = 2D$, and $\rho' = \rho/2$, the new frequency $n' = \frac{1}{2D} \sqrt{\frac{2T}{\pi (\rho/2)}} = \frac{1}{2D} \sqrt{\frac{4T}{\pi \rho}} = n$.
A string is cut into three parts, having fundamental frequencies $n_1$, $n_2$ and $n_3$ respectively. Then original fundamental frequency ‘$n$’ related by the expression as:
(2000)
For a given string, fundamental frequency $n \propto 1/l$. For the cut segments, $l = l_1 + l_2 + l_3$. Substituting $l = k/n$, we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.
A cylindrical tube ($L = 125 \text{ cm}$) is resonant with a tuning fork of frequency $330 \text{ Hz}$. If it is filling by water then to get resonance again, minimum length of water column is ($v = 330 \text{ m/s}$):
(1999)
Wavelength $\lambda = \frac{v}{f} = \frac{330}{330} = 1 \text{ m} = 100 \text{ cm}$. Resonance occurs at air column lengths $L_{air} = \lambda/4, 3\lambda/4, 5\lambda/4... = 25 \text{ cm}, 75 \text{ cm}, 125 \text{ cm}$. With water filling, to find minimum water length, we need the maximum resonant air length less than the tube length, which is $75 \text{ cm}$. Minimum water length = $125 - 75 = 50 \text{ cm}$.
A standing wave having 3 nodes and 2 antinodes is formed between $1.21 \text{ \AA}$ distance then the wavelength is:
(1998)
A standing wave with 3 nodes and 2 antinodes corresponds to 2 full loops. The length of one loop is $\lambda/2$. Total distance $L = 2 \times (\lambda/2) = \lambda$. Therefore, the wavelength $\lambda = 1.21 \text{ \AA}$.
The length of a sonometer wire AB is $110 \text{ cm}$. Where should the two bridges be placed from A to divide the wire in 3 segments whose fundamental frequencies are in the ratio of $1:2:3$?
(1995)
Since frequency $f \propto 1/l$, the lengths will be in the ratio $1/1 : 1/2 : 1/3 = 6:3:2$. The sum of the ratios is 11. Lengths are $60 \text{ cm}$, $30 \text{ cm}$, and $20 \text{ cm}$. The bridges should be placed at $60 \text{ cm}$ and $60 + 30 = 90 \text{ cm}$ from A.