Grade 11 : Physics - Waves (Set 3)

Explanation:Comparing with standard equation $y(x,t) = A \sin(kx - \omega t)$, we identify $k = 4\pi\text{ rad/m}$ and $\omega = 200\pi\text{ rad/s}$. The wave velocity is $v = \frac{\omega}{k} = \frac{200\pi}{4\pi} = 50m/s$. The term $(kx - \omega t)$ signifies propagation along the positive $x$-axis.

Explanation:Linear mass density of string is $\mu = \frac{m}{L} = \frac{0.16kg}{4m} = 0.04\text{ kg/m}$. Wave speed on a stretched string is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.04}} = \sqrt{2500} = 50m/s$.

Explanation:Path difference $\Delta x = d_2 - d_1 = 3.60m - 3.45m = 0.15m$. Phase difference is $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{0.60} \times 0.15 = \frac{\pi}{2}\text{ rad}$.

Explanation:For a string fixed at both ends, frequencies are given by $f_n = n f_1$, where $n = 1, 2, 3...$. The second overtone corresponds to the third harmonic ($n = 3$). Thus, $f_3 = 3 \times 150\text{ Hz} = 450\text{ Hz}$.

Explanation:Fundamental frequency of open pipe is $f_{\text{open}} = \frac{v}{2 L_1}$. Fundamental frequency of closed pipe is $f_{\text{closed}} = \frac{v}{4 L_2}$. Setting them equal: $\frac{v}{2 L_1} = \frac{v}{4 L_2} \implies \frac{L_1}{L_2} = \frac{4}{2} = 2$. Ratio $L_1 : L_2 = 2 : 1$.

Explanation:Initial frequency $f_B$ could be $256 \pm 4 = 260\text{ Hz}$ or 252 Hz. Loading wax reduces the frequency of B ($f_B'$ drops). If $f_B = 260\text{ Hz}$, dropping $f_B$ brings it closer to 256 Hz, decreasing beats to 2. Thus, original frequency of B was 260 Hz.

Explanation:Doppler effect for source moving towards stationary observer: $f' = f_0 \left(\frac{v}{v - v_s}\right) = 600 \left(\frac{340}{340 - 34}\right) = 600 \left(\frac{340}{306}\right) = 666.67\text{ Hz}$.

Explanation:Speed of sound in gas is $v = \sqrt{\frac{\gamma R T}{M}}$. Since $\gamma, R, T$ are identical, $v \propto \frac{1}{\sqrt{M}}$. Therefore, $\frac{v_{O2}}{v_{H2}} = \sqrt{\frac{M_{H2}}{M_{O2}}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.

Explanation:Maximum particle velocity $v_{p,\max} = A\omega$. Wave velocity $v_w = \frac{\omega}{k}$. Given $A\omega = 3 \left(\frac{\omega}{k}\right) \implies A k = 3$. Substituting $k = \frac{2\pi}{\lambda}$, we get $A \left(\frac{2\pi}{\lambda}\right) = 3 \implies \lambda = \frac{2\pi A}{3}$.

Explanation:Fundamental frequency of closed pipe is $f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.85} = 100\text{ Hz}$. A pipe closed at one end produces only odd harmonics ($f_1, f_3, f_5...$). Third harmonic is $f_3 = 3 f_1 = 300\text{ Hz}$.

Explanation:New intensity $I' \propto (f/2)^2 (2A)^2 = \frac{f^2}{4} \times 4 A^2 = f^2 A^2$. The factor of $1/4$ from frequency and factor of 4 from amplitude cancel out, so intensity remains unchanged.

Explanation:For an open organ pipe (open at both ends), end correction $e$ applies at both ends. $L_{\text{eff}} = L + 2e = 30.0 + 2(0.60 \times 1.0) = 30.0 + 1.2 = 31.2cm$.

Explanation:Phase difference $\phi = \pi/2$. Resultant amplitude is $A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} = \sqrt{3^2 + 4^2 + 2(3)(4)\cos(\pi/2)} = \sqrt{9 + 16 + 0} = 5\text{ units}$.

Explanation:Fundamental frequency $f \propto \sqrt{T}$. New tension $T' = 1.44 T$. New frequency $f' = f \sqrt{1.44} = 250 \times 1.2 = 300\text{ Hz}$.

Explanation:Reflection at a rigid boundary introduces a phase shift of $\pi\text{ rad}$ and reverses propagation direction ($x \to -x$). Thus $y_r(x,t) = 0.02 \sin(20\pi t + 4\pi x + \pi) = -0.02 \sin(20\pi t + 4\pi x)$.

Explanation:Wavelength $\lambda = \frac{v}{f} = \frac{320}{400} = 0.80m$. Distance between node and adjacent antinode is $\frac{\lambda}{4} = \frac{0.80}{4} = 0.20m$.

Explanation:Doppler effect for observer moving towards stationary source: $f' = f_0 \left(\frac{v + v_o}{v}\right) = 500 \left(\frac{340 + 20}{340}\right) = 500 \times \frac{360}{340} = 529.41\text{ Hz}$.

Explanation:Speed of sound $v \propto \sqrt{T\text{ (in K)}}$. Thus $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = 2 \implies T_2 = 4 T_1$. Given $T_1 = 273\text{ K}$, $T_2 = 4 \times 273 = 1092\text{ K}$. In Celsius: $t_2 = 1092 - 273 = 819^\circ\text{C}$.

Explanation:Resonance positions satisfy $L_1 + e = \frac{\lambda}{4}$ and $L_2 + e = \frac{3\lambda}{4}$. Subtracting gives $L_2 - L_1 = \frac{\lambda}{2} \implies \lambda = 2(L_2 - L_1) = 2(50 - 16) = 68cm$.

Explanation:Substitute $x = 0.25m$ and $t = 0.10\text{ s}$ into the wave function: $y = 0.04 \cos(10\pi (0.10) - 2\pi (0.25)) = 0.04 \cos(\pi - 0.5\pi) = 0.04 \cos(0.5\pi) = 0m$.