Sound Waves — NEET Physics PYQ MCQs with Solutions

Solution: Velocity of a wave is given by $v = n\lambda$ where, $n$ = frequency of wave, $\lambda$ = wavelength of wave So, for two different cases: $v_1 = n_1\lambda_1$ $v_2 = n_2\lambda_2$ $\frac{\lambda_2}{\lambda_1} = \frac{\lambda_1 v_2}{v_1} = \lambda_1 \times \frac{3500}{350} = \lambda_1 \times 10$ [Since $n_1 = n_2$] Therefore, $\lambda_2 = 10\lambda_1$

Solution: Sabine's formula for reverberation time is $T = 0.16\frac{V}{\sum as}$ where, V is volume of hall in m³. $\sum as$ = total absorption of the hall (room) So, $T \propto \frac{V}{s}$ For two different cases of reverberation: $\frac{T'}{T} = \frac{V'}{V} \times \frac{s}{s'} = \frac{(2)^3}{(2)^2} = \frac{8}{4} = 2$ Hence, $T' = 2T = 2 \times 1 = 2$ s

Solution: Intensity of sound, $I = \frac{p}{4\pi r^2}$ or $I \propto \frac{1}{r^2}$. Therefore, $\frac{I_1}{I_2} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Solution: The presence of water vapours reduces the density of air, which increases the velocity of sound: $v_m = \sqrt{\frac{\rho_d}{\rho_m}} v_d > v_d$ since $\rho_m < \rho_d$

Solution: Wavelength = $\frac{v}{f} = \frac{1.7 \times 10^3}{4.2 \times 10^6} = 4 \times 10^{-4}$ m

Solution: Velocity of sound: $c_s = \sqrt{\frac{\gamma p}{\rho}}$. RMS velocity: $c = \sqrt{\frac{3p}{\rho}}$. Therefore, $c_s = \frac{c}{\sqrt{3}}$

Solution: Intensity $I \propto A^2 \times f^2$. New intensity = $(2A)^2 \times (f/4)^2 = 4A^2 \times f^2/16 = \frac{1}{4} A^2 f^2$. Intensity decreases by a factor of 4.

Solution: Velocity of sound: $v = \sqrt{\frac{E}{\rho}}$ where E is elasticity and $\rho$ is density of the medium.

Solution: Fundamental frequency $f \propto \frac{1}{l}$. So $\frac{f_1}{f_2} = \frac{l_2}{l_1}$. $\frac{120}{180} = \frac{l_2}{90}$, giving $l_2 = 60$ cm

Solution: Beat frequency = $|f_A - f_B|$. Initially $f_B = 530 \pm 6$ Hz = 536 or 524 Hz. When tension decreases, frequency decreases. If $f_B = 536$ Hz, beat frequency would decrease, not increase. So $f_B = 524$ Hz.

Solution: For closed pipe: $l_1 = \frac{\lambda}{4}$, $l_2 = \frac{3\lambda}{4}$. So $l_2 - l_1 = \frac{\lambda}{2}$. $\lambda = 2(31.25 - 9.75) = 43$ cm. $v = f \times \lambda = 800 \times 0.43 = 344$ m/s

Solution: For open pipe: $f_1 = \frac{v}{2L}$. For closed pipe, 3rd harmonic: $f_3 = \frac{3v}{4L_c}$. Setting equal: $\frac{v}{2L} = \frac{3v}{4 \times 20}$, so $L = 13.3$ cm

Solution: Successive resonances differ by $\frac{\lambda}{2}$. $l_2 - l_1 = 73 - 20 = 53$ cm = $\frac{\lambda}{2}$. $\lambda = 1.06$ m. $v = f \times \lambda = 320 \times 1.06 = 339.2$ m/s

Solution: In a closed pipe, only odd harmonics exist. Consecutive odd harmonics differ by 2 orders. Frequency difference = $\frac{2v}{4l}$. $260 - 220 = 40 = 2 \times \frac{v}{4l}$. So $\frac{v}{4l} = 20$ Hz (fundamental)

Solution: Second overtone of open pipe (3rd harmonic): $f = \frac{3v}{2L_o}$. First overtone of closed pipe (3rd harmonic): $f = \frac{3v}{4L}$. Setting equal: $\frac{3v}{2L_o} = \frac{3v}{4L}$, so $L_o = 2L$

Solution: Beat frequency between (n-1) and n: $n - (n-1) = 1$ Hz. Beat frequency between n and (n+1): $(n+1) - n = 1$ Hz. Total beats = 1 Hz

Solution: Fundamental of closed pipe: $f_c = \frac{v}{4l_c}$. Second overtone (3rd harmonic) of open pipe: $f_o = \frac{3v}{2l_o}$. Setting equal: $\frac{v}{4 \times 0.2} = \frac{3v}{2l_o}$, so $l_o = 1.2$ m = 120 cm

Solution: Smallest resonant length: $L_{min} = \frac{\lambda}{4} = 50$ cm, so $\lambda = 200$ cm. Next resonant length: $L = \frac{3\lambda}{4} = \frac{3 \times 200}{4} = 150$ cm

Solution: For pipe closed at one end: $f_n = \frac{nv}{4l}$ where $n$ is odd. Here, $f_n = \frac{n \times 340}{4 \times 85 \times 10^{-2}} = 100n$ Hz. For $f_n &lt; 1250$ Hz: $100n &lt; 1250 \Rightarrow n &lt; 12.5$. Since $n$ is odd, $n = 1, 3, 5, 7, 9, 11$, giving 6 possible oscillations.

Solution: At open ends, both displacement antinode and pressure node occur. Therefore, pressure change is zero at both open ends, not maximum. Statement (d) is not true.

Solution: From law of length: $n \propto \frac{1}{l}$, so $nl = k$ (constant). For segments: $l_1 = \frac{k}{n_1}$, $l_2 = \frac{k}{n_2}$, etc. Total length: $l = l_1 + l_2 + l_3 + ... = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3} + ...$. Since $l = \frac{k}{n}$: $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} + ...$

Solution: With 3 nodes and 2 antinodes, there are 2 complete segments. Each segment = $\frac{\lambda}{2}$. Total length = $2 \times \frac{\lambda}{2} = \lambda = 1.21$ Å

Solution: Original (open pipe): $f = \frac{v}{2l}$. After dipping half length: new air column length = $\frac{l}{2}$, pipe becomes closed-open (like closed pipe). New frequency: $f' = \frac{v}{4 \times l/2} = \frac{v}{2l} = f$. No change in fundamental frequency.

Solution: At a fixed end, a pulse undergoes reflection with a phase change of $\pi$ (180°). The wave velocity reverses direction after reflection.

Solution: With 5 segments: $5 \times \frac{\lambda}{2} = 10$ m, so $\lambda = 4$ m. Frequency: $n = \frac{v}{\lambda} = \frac{20}{4} = 5$ Hz

Solution: Distance between successive nodes = $\frac{\lambda}{2}$. Since $v = n\lambda$: $\lambda = \frac{v}{n} = \frac{20}{n}$. Distance between nodes = $\frac{\lambda}{2} = \frac{10}{n}$

Solution: At fixed end, a node always forms. Distance between consecutive nodes = $\frac{\lambda}{2} = 10$ cm, so $\lambda = 20$ cm = 0.2 m. Velocity: $v = n\lambda = 100 \times 0.2 = 20$ m/s

Solution: Standard standing wave form: $y = a\sin(\omega t)\cos(kx)$. Here: $\omega = 100$, $k = 0.01$. Wave velocity: $v = \frac{\omega}{k} = \frac{100}{0.01} = 10^4$ m/s

Solution: Since $\nu \propto \frac{1}{L}$: $\frac{\nu_1}{\nu_2} = \frac{L_2}{L_1}$. $\frac{512}{256} = \frac{L_2}{0.5}$, so $L_2 = 2 \times 0.5 = 1$ m

Solution: Third overtone is the 7th harmonic (7 = 2×4-1, where 4 is mode number). For closed pipe in 7th harmonic: $L = \frac{7\lambda}{4} = \frac{\lambda}{2} + \frac{\lambda}{2} + \frac{\lambda}{2} + \frac{\lambda}{4}$. This gives 4 complete nodes and 4 antinodes.