In the resonance experiment, two air columns (closed at one end) of 100mathrm~cm$100\mathrm{~cm}$ and 120mathrm~cm$120\mathrm{~cm}$ long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :

### Related Formula v = sqrtfracTmu$$v = \sqrt{\frac{T}{\mu}}$$ t = fracLv$$t = \frac{L}{v}$$ ### Core Logic Given L_A = 2.5text m$L_{A} = 2.5\text{ m}$, L_B = 1.5text m$L_{B} = 1.5\text{ m}$, T = 500text N$T = 500\text{ N}$. Velocity in string A: v_A = sqrtfracTmu_A = sqrtfrac5002 times 10^-4 = sqrt2500000 = 5 sqrt10 times 10^2text m/s$$v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = \sqrt{2500000} = 5 \sqrt{10} \times 10^{2}\text{ m/s}$$ Velocity in string B: v_B = sqrtfracTmu_B = sqrtfrac5004 times 10^-4 = sqrt1250000 = 5 sqrt5 times 10^2text m/s$$v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = \sqrt{1250000} = 5 \sqrt{5} \times 10^{2}\text{ m/s}$$ ### Step 1: Calculate Ratio of Times Time taken to reach joint: t_1 = fracL_Av_A = frac2.55 sqrt10 times 10^2$$t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5 \sqrt{10} \times 10^{2}}$$ t_2 = fracL_Bv_B = frac1.55 sqrt5 times 10^2$$t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5 \sqrt{5} \times 10^{2}}$$ Ratio: fract_1t_2 = frac2.55sqrt10 times frac5sqrt51.5 = frac2.51.5 times fracsqrt5sqrt10 = frac53 times frac1sqrt2$$\frac{t_{1}}{t_{2}} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{2.5}{1.5} \times \frac{\sqrt{5}}{\sqrt{10}} = \frac{5}{3} \times \frac{1}{\sqrt{2}}$$ fract_1t_2 = frac1.6661.414 approx 1.18$$\frac{t_{1}}{t_{2}} = \frac{1.666}{1.414} \approx 1.18$$ ### Pattern Recognition Time ratio is proportional to fracLsqrtT/mu = L sqrtmu/T$\frac{L}{\sqrt{T/\mu}} = L \sqrt{\mu/T}$. Thus t_1/t_2 = (L_1/L_2) sqrtmu_1/mu_2$t_1/t_2 = (L_1/L_2) \sqrt{\mu_1/\mu_2}$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula 1. Wave function (moving along +x$+x$ direction) with a peak at x=0, t=0$x=0, t=0$: y(x, t) = A cos(kx - omega t)$$y(x, t) = A \cos(kx - \omega t)$$ 2. Wave number: k = frac2pilambda$$k = \frac{2\pi}{\lambda}$$ 3. Wave speed: v = fracomegak$$v = \frac{\omega}{k}$$ ### Core Logic Given parameters: - Wavelength lambda = 7.5 \ mathrmcm$\lambda = 7.5 \ \mathrm{cm}$ - Distance travelled Delta x = 1.2 \ mathrmcm$\Delta x = 1.2 \ \mathrm{cm}$ in Delta t = 0.3 \ mathrms$\Delta t = 0.3 \ \mathrm{s}$ - Maximum displacement (amplitude) A = 2 \ mathrmcm$A = 2 \ \mathrm{cm}$ Let's calculate the wave parameters: 1. **Wave number (k$k$):** k = frac2pi7.5 = frac20pi75 = frac4pi15 approx 0.838 \ mathrmrad/cm$$k = \frac{2\pi}{7.5} = \frac{20\pi}{75} = \frac{4\pi}{15} \approx 0.838 \ \mathrm{rad/cm}$$ 2. **Wave speed (v$v$):** v = fracDelta xDelta t = frac1.20.3 = 4 \ mathrmcm/s$$v = \frac{\Delta x}{\Delta t} = \frac{1.2}{0.3} = 4 \ \mathrm{cm/s}$$ 3. **Angular frequency (omega$\omega$):** omega = v cdot k = 4 times frac4pi15 = frac16pi15 approx 3.35 \ mathrmrad/s$$\omega = v \cdot k = 4 \times \frac{4\pi}{15} = \frac{16\pi}{15} \approx 3.35 \ \mathrm{rad/s}$$ Since the crest is at x=0$x=0$ at t=0$t=0$, y(0,0) = 2 = A$y(0,0) = 2 = A$. This boundary condition demands a cosine function. ### Step 1: Write wave equation Substitute A$A$, k$k$, and omega$\omega$ into the standard form: y(x, t) = 2 cos(0.83x - 3.35t) \ mathrmcm$$y(x, t) = 2 \cos(0.83x - 3.35t) \ \mathrm{cm}$$ This perfectly matches Option (1). ### Pattern Recognition Sees: Wavelength and speed to determine travelling wave equation. Trap: Choosing sine instead of cosine. Since the crest (maximum displacement) is at x=0, t=0$x=0, t=0$, y(0,0)$y(0,0)$ must equal A$A$, which is satisfied only by the cosine function. Shortcut: Calculate k = 2pi / 7.5 approx 0.83$k = 2\pi / 7.5 \approx 0.83$. This immediately eliminates Options (3) and (4). Calculate speed v = 4$v = 4$, so omega = 4 times 0.83 approx 3.35$\omega = 4 \times 0.83 \approx 3.35$, which points directly to Option (1). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula The product of cosines can be transformed into a sum using the trigonometric identity: cos A cos B = frac12 [cos(A + B) + cos(A - B)]$$\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$$ The beat frequency f_textbeat$f_{\text{beat}}$ is given by: f_textbeat = |f_1 - f_2| = left| fracomega_1 - omega_22pi right|$$f_{\text{beat}} = |f_1 - f_2| = \left| \frac{\omega_1 - \omega_2}{2\pi} \right|$$ The beat period T_textbeat$T_{\text{beat}}$ is: T_textbeat = frac1f_textbeat$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}}$$ ### Core Logic Rewrite the superposition equation: x = a cos(1.5t) cos(50.5t)$$x = a \cos(1.5t) \cos(50.5t)$$ Apply the identity with A = 50.5t$A = 50.5t$ and B = 1.5t$B = 1.5t$: x = fraca2 [cos(52t) + cos(49t)]$$x = \frac{a}{2} [\cos(52t) + \cos(49t)]$$ Here, the two component frequencies are: omega_1 = 52 mathrm~rad/s implies f_1 = frac522pi$$\omega_1 = 52 \mathrm{~rad/s} \implies f_1 = \frac{52}{2\pi}$$ omega_2 = 49 mathrm~rad/s implies f_2 = frac492pi$$\omega_2 = 49 \mathrm{~rad/s} \implies f_2 = \frac{49}{2\pi}$$ Calculate the beat frequency: f_textbeat = f_1 - f_2 = frac52 - 492pi = frac32pi mathrm~Hz$$f_{\text{beat}} = f_1 - f_2 = \frac{52 - 49}{2\pi} = \frac{3}{2\pi} \mathrm{~Hz}$$ ### Step 1: Calculate Beat Period The time period of beats is: T_textbeat = frac1f_textbeat = frac2pi3 approx frac2 times 3.143 = 2.09 mathrm~s$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3} \approx \frac{2 \times 3.14}{3} = 2.09 \mathrm{~s}$$ Rounding to the nearest integer gives 2 mathrm~s$2 \mathrm{~s}$. ### Pattern Recognition Sees: product of two cosines with significantly different coefficients omega_1$\omega_1$ and omega_2$\omega_2$. Shortcut: The beat period is simply 2pi$2\pi$ divided by the difference between the two component frequencies, where the component frequencies are (omega_textaverage pm omega_textenvelope)$(\omega_{\text{average}} \pm \omega_{\text{envelope}})$. The difference is 2 times omega_textenvelope = 2 times 1.5 = 3 mathrm~rad/s$2 \times \omega_{\text{envelope}} = 2 \times 1.5 = 3 \mathrm{~rad/s}$. Thus, T = 2pi / 3 approx 2 mathrm~s$T = 2\pi / 3 \approx 2 \mathrm{~s}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

### Related Formula A_textres = sqrtA_1^2 + A_2^2 + 2 A_1 A_2 cosphi$$A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}$$ tantheta = fracA_2 sinphiA_1 + A_2 cosphi$$\tan\theta = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi}$$ where, A_1, A_2$A_1, A_2$ = amplitudes of individual harmonic waves phi$\phi$ = phase difference between the waves A_textres$A_{\text{res}}$ = resultant amplitude theta$\theta$ = resultant phase angle relative to the first wave ### Core Logic Given parameters: - A_1 = 4$A_1 = 4$ - A_2 = 2$A_2 = 2$ - Phase difference, phi = frac2pi3 = 120^circ$\phi = \frac{2\pi}{3} = 120^\circ$

Calculate Resultant Amplitude: A_textres = sqrt4^2 + 2^2 + 2(4)(2) cos 120^circ$$A_{\text{res}} = \sqrt{4^2 + 2^2 + 2(4)(2) \cos 120^\circ}$$ A_textres = sqrt16 + 4 + 16 left(-0.5right) = sqrt20 - 8 = sqrt12 = 2sqrt3$$A_{\text{res}} = \sqrt{16 + 4 + 16 \left(-0.5\right)} = \sqrt{20 - 8} = \sqrt{12} = 2\sqrt{3}$$ Calculate Resultant Phase (angle theta$\theta$): tantheta = frac2 sin 120^circ4 + 2 cos 120^circ = frac2 left(fracsqrt32right)4 + 2 left(-0.5right) = fracsqrt33 = frac1sqrt3$$\tan\theta = \frac{2 \sin 120^\circ}{4 + 2 \cos 120^\circ} = \frac{2 \left(\frac{\sqrt{3}}{2}\right)}{4 + 2 \left(-0.5\right)} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$ theta = fracpi6$$\theta = \frac{\pi}{6}$$ ### Step 1: Result Format Represent the resultant amplitude and phase as a pair: left[2sqrt3, fracpi6right]$$\left[2\sqrt{3}, \frac{\pi}{6}\right]$$ ### Pattern Recognition Sees: Vector-like addition of wave amplitudes. Shortcut: Solve it like a vector addition problem. Vector A_1$A_1$ along horizontal (0$0$), and Vector A_2$A_2$ at 120^circ$120^\circ$. The magnitude is sqrt4^2+2^2-2(4)(2)(0.5) = 2sqrt3$\sqrt{4^2+2^2-2(4)(2)(0.5)} = 2\sqrt{3}$ and direction is pi/6$\pi/6$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves