Question 1:
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 2:
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 3:
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 4:
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 5:
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 6:
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 7:
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 $I_{max}/I_{min} = (A_1 + A_2)^2 / (A_1 - A_2)^2 = (5+4)^2 / (5-4)^2 = 81:1$.
Question 8:
easyTwo identical piano wires, kept under the same tension $T$ have a fundamental frequency of $600 \text{ Hz}$. The fractional increase in the tension of one of the wires which will lead to occurrence of $6 \text{ beats/s}$ when both the wires oscillate together would be:
(2011 Mains)
Fundamental frequency $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$. Fractional change is $$\frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T}$$. Given $\Delta f = 6$ and $f = 600$, we have $$\frac{\Delta T}{T} = 2 \times \frac{6}{600} = 0.02$$.
Question 9:
moderateA tuning fork of frequency $512 \text{ Hz}$ makes 4 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was:
(2010 Pre)
Possible frequency of string is $512 \pm 4 = 508$ or $516 \text{ Hz}$. Increasing tension increases frequency. If it were $516$, new frequency would be $>516$, increasing beats $>4$. If it was $508$, new frequency could be $510$, decreasing beats to 2. Original was $508 \text{ Hz}$.
Question 10:
moderateEach of the two strings of length $51.6 \text{ cm}$ and $49.1 \text{ cm}$ are tensioned separately by $20 \text{ N}$ force. Mass per unit length of both the strings is same and equal to $1 \text{ g/m}$. When both the strings vibrate simultaneously the number of beats is:
(2009)
Velocity $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = 141.42 \text{ m/s}$. The frequencies are $f_1 = \frac{v}{2L_1} = \frac{141.42}{2 \times 0.516} \approx 137 \text{ Hz}$ and $f_2 = \frac{141.42}{2 \times 0.491} \approx 144 \text{ Hz}$. Number of beats = $144 - 137 = 7$.