Question 11:
easyAssertion (A): Interference can happen in sound waves.
Reason (R): In Quincke’s tube, interference is present due to initial phase difference as well as the phase difference due to path difference.
Sound waves, being waves, exhibit interference. Thus, Assertion (A) is true. In Quincke's tube, interference is primarily due to path difference. An initial phase difference is not typically considered in a standard setup. Therefore, Reason (R) is false.
Question 12:
easyAssertion (A): It is not possible to have interference between the waves produced by two violins of different frequency.
Reason (R): For interference of two waves, the phase difference between the waves must remain constant.
Assertion (A) is true. For sustained interference, sources must be coherent, meaning same frequency and constant phase difference. Two violins produce incoherent waves. Reason (R) is true and correctly states the condition for interference.
Question 13:
easyAssertion (A): Energy is created during constructive interference and destroyed during destructive interference.
Reason (R): The positions of constructive interference are sources of energy while the positions of destructive interference are sinks of energy.
Energy is always conserved in interference phenomena; it is merely redistributed, not created or destroyed. Thus, both Assertion (A) and Reason (R) are false.
Question 14:
moderateIf the amplitude of sound is doubled and the frequency reduced to one fourth, the intensity of sound at the same point will be:
(1989)
Intensity $I \propto A^2 f^2$. New intensity $I' \propto (2A)^2 (f/4)^2 = 4A^2 \times \frac{f^2}{16} = \frac{1}{4} I$. Thus it decreases by a factor of 4.
Question 15:
moderateTwo periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
(2008)
$I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2$. Sum $= (I_1 + I_2 + 2\sqrt{I_1 I_2}) + (I_1 + I_2 - 2\sqrt{I_1 I_2}) = 2(I_1 + I_2)$.
Question 16:
moderateA 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 = $125 - 75 = 50 \text{ cm}$.
Question 17:
easyIn a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6 \text{ Hz}$. When tension in B is slightly decreased, the beat frequency increases to $7 \text{ Hz}$. If the frequency of A is $530 \text{ Hz}$, the original frequency of B will be :
(2020)
Frequency of string B is $f_B = f_A \pm 6 = 530 \pm 6$. It can be $536 \text{ Hz}$ or $524 \text{ Hz}$. Decreasing tension lowers frequency. If $f_B = 524$, lowering it increases the difference from 530, producing a higher beat frequency ($7 \text{ Hz}$). Thus, original $f_B$ was $524 \text{ Hz}$.
Question 18:
moderateThree sound waves of equal amplitudes have frequencies $(n – 1), n, (n + 1)$. They superimpose to give beats. The number of beats produced per second will be:
(2016 – II)
The number of beats is determined by the maximum frequency difference among the superimposing waves. Max beat frequency = $(n+1) - (n-1) = 2$.
Question 19:
moderateA source of unknown frequency gives $4 \text{ beats/s}$, when sounded with a source of known frequency $250 \text{ Hz}$. The second harmonic of the source of unknown frequency gives five beats per second, when sounded with a source of frequency $513 \text{ Hz}$. The unknown frequency is:
(2013)
Unknown frequency $f = 250 \pm 4 = 254 \text{ Hz}$ or $246 \text{ Hz}$. For the second harmonic, $2f$ must give 5 beats with 513 Hz. If $f = 254$, $2f = 508 \implies |513 - 508| = 5$. If $f = 246$, $2f = 492 \implies |513 - 492| = 21$. So $f = 254 \text{ Hz}$.
Question 20:
easyTwo sources of sound placed close to each other are emitting progressive waves given by $y_1 = 4 \sin 600 \pi t$ and $y_2 = 5 \sin 608 \pi t$. An observer located near these two sources of sound will hear:
(2012 Pre)
From the equations, $\omega_1 = 600\pi \implies f_1 = 300 \text{ Hz}$, and $\omega_2 = 608\pi \implies f_2 = 304 \text{ Hz}$. Beat frequency = $304 - 300 = 4 \text{ Hz}$. Intensity ratio $I_{max}/I_{min} = (A_1 + A_2)^2 / (A_1 - A_2)^2 = (5+4)^2 / (5-4)^2 = 81:1$.