Question 21:
moderateEquation of progressive wave is given by $y = 4 \sin[\pi(\frac{t}{5} – \frac{x}{9}) + \frac{\pi}{6}]$ where $y, x$ are in cm and $t$ is in seconds. Then which of the following is correct?
(1988)
Comparing with $y = A \sin(2\pi(\frac{t}{T} - \frac{x}{\lambda}) + \phi)$, rewrite the given equation as $y = 4 \sin[2\pi(\frac{t}{10} - \frac{x}{18}) + \frac{\pi}{6}]$. This gives $\lambda = 18 \text{ cm}$.
Question 22:
moderateA wave travelling in positive $x$-direction with $A = 0.2 \text{ m}$ velocity $= 360 \text{ m/s}$ and $\lambda = 60 \text{ m}$, then correct expression for the wave is:
(2002)
Frequency $n = \frac{v}{\lambda} = \frac{360}{60} = 6 \text{ Hz}$. Equation for wave in positive x-direction is $y = A \sin[2\pi(nt - \frac{x}{\lambda})] = 0.2 \sin[2\pi(6t - \frac{x}{60})]$.
Question 23:
easyThe equation of a wave is represented by $y = 10^{-4} \sin(100t – \frac{x}{10}) \text{ m}$, then the velocity of wave will be:
(2001)
Comparing with $y = a \sin(\omega t - kx)$, we get $\omega = 100$ and $k = 1/10$. Velocity $v = \frac{\omega}{k} = \frac{100}{1/10} = 1000 \text{ m/s}$.
Question 24:
easyFor a wave $y = y_0 \sin(\omega t – kx)$, for what value of $\lambda$ is the maximum particle velocity equal to two times the wave velocity: (1998)
Maximum particle velocity $v_{max} = A\omega = y_0\omega$. Wave velocity $v = \frac{\omega}{k} = \frac{\omega\lambda}{2\pi}$. Given $y_0\omega = 2(\frac{\omega\lambda}{2\pi})$, resolving gives $\lambda = \pi y_0$.
Question 25:
The equation of a sound wave is $y = 0.0015 \sin(62.4x + 316t)$. The wavelength of this wave is:
(1996)
Here $k = 62.4$. Since $k = \frac{2\pi}{\lambda}$, we have $\lambda = \frac{2 \times 3.14}{62.4} \approx 0.1 \text{ unit}$.
Question 26:
Two sound waves having a phase difference of $60^{\circ}$ have path difference of:
(1996)
Phase difference $\Delta\phi = \frac{2\pi}{\lambda} \times \Delta x$. Given $\Delta\phi = 60^{\circ} = \frac{\pi}{3}$. Thus, $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x \implies \Delta x = \frac{\lambda}{6}$.
Question 27:
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 28:
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)
$v = 2f(L_2 - L_1) = 2 \times 320 \times (0.73 - 0.20) = 640 \times 0.53 = 339.2 \text{ m/s}$.
Question 29:
moderateThe 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, $f_0 = 20 \text{ Hz}$.
Question 30:
easyThe 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: $3v / 2L_{\text{open}} = 3v / 4L \Rightarrow L_{\text{open}} = 2L$.