Question 41:
moderateA stationary wave is represented by $y = A \sin(100t) \cos(0.01x)$, where $y$ and $A$ are in millimetres, $t$ is in seconds and $x$ is in metres. The velocity of the wave is:
(1994)
For the constituent waves, $\omega = 100$ and $k = 0.01$. Velocity $v = \frac{\omega}{k} = \frac{100}{0.01} = 10^4 \text{ m/s}$.
Question 42:
moderateA wave of frequency $100 \text{ Hz}$ travels along a string towards its fixed end. When this wave travels back, after reflection, a node is formed at a distance of $10 \text{ cm}$ from the fixed end. The speed of the wave (incident and reflected) is:
(1994)
Distance from fixed end (node) to nearest node is $\frac{\lambda}{2} = 10 \text{ cm} = 0.1 \text{ m}$. Thus $\lambda = 0.2 \text{ m}$. Velocity $v = f\lambda = 100 \times 0.2 = 20 \text{ m/s}$.
Question 43:
moderateAn 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:
(2016-I)
Smallest length $L_1 = \lambda/4 = 50 \text{ cm}$. Next resonating length $L_2 = 3\lambda/4 = 3 L_1 = 3 \times 50 = 150 \text{ cm}$.
Question 44:
moderateThe fundamental frequency of a closed organ pipe of length $20 \text{ cm}$ is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is:
(2015)
Fundamental of closed = $v / (4 \times 20)$. 2nd overtone of open = $3v / 2L$. So, $v / 80 = 3v / 2L \Rightarrow 2L = 240 \Rightarrow L = 120 \text{ cm}$.
Question 45:
easyA string is stretched between fixed points separated by $75.0 \text{ cm}$. It is observed to have resonant frequencies of $420 \text{ Hz}$ and $315 \text{ Hz}$. There are no other resonant frequencies between these two. The lowest resonant frequencies for this string is:
(2015 Re)
The lowest resonant (fundamental) frequency $f$ is the difference between two consecutive resonant frequencies. $f = f_{n+1} - f_n = 420 - 315 = 105 \text{ Hz}$.
Question 46:
moderateThe number of possible natural oscillations of air column in a pipe closed at one end of length $85 \text{ cm}$ whose frequencies lie below $1250 \text{ Hz}$ are (velocity of sound = $340 \text{ m s}^{-1}$):
(2014)
Fundamental $f_0 = v / 4L = 340 / (4 \times 0.85) = 100 \text{ Hz}$. Frequencies are odd multiples: 100, 300, 500, 700, 900, 1100. All lie below 1250 Hz. Count is 6.
Question 47:
moderateIf $n_1, n_2$ and $n_3$ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency $n$ of the string is given by:
(2014)
Total length $L = L_1 + L_2 + L_3$. Since frequency $n \propto 1/L \Rightarrow L \propto 1/n$. Substituting lengths, we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.
Question 48:
easyThe length of the string of a musical instrument is $90 \text{ cm}$ and has a fundamental frequency of $120 \text{ Hz}$. Where should it be pressed to produce fundamental frequency of $180 \text{ Hz}$?
(2020-Covid)
Fundamental frequency $f \propto \frac{1}{L}$. So $f_1 L_1 = f_2 L_2$. $120 \times 90 = 180 \times L_2 \Rightarrow L_2 = 60 \text{ cm}$.
Question 49:
moderateThe fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \text{ cm}$, the length of the open organ pipe is:
(2018)
Fundamental of open = $v / 2L_o$. Third harmonic of closed = $3v / 4L_c$. Equating them: $v / 2L_o = 3v / 4L_c \Rightarrow L_o = 2L_c / 3 = 2(20)/3 = 13.33 \text{ cm} \approx 13.2 \text{ cm}$.
Question 50:
difficultA 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:
(2018)
$v = 2f(L_2 - L_1) = 2 \times 320 \times (0.73 - 0.20) = 640 \times 0.53 = 339.2 \text{ m/s}$.