### 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 For an air column closed at one end, the fundamental frequency f$f$ is given by: f = fracv4l$f = \frac{v}{4l}$ where v$v$ is the velocity of sound and l$l$ is the length of the air column. ### Core Logic Given parameters: - l_1 = 100mathrm~cm = 1.0mathrm~m$l_1 = 100\mathrm{~cm} = 1.0\mathrm{~m}$ - l_2 = 120mathrm~cm = 1.2mathrm~m$l_2 = 120\mathrm{~cm} = 1.2\mathrm{~m}$ - Beats per second (f_1 - f_2$f_1 - f_2$) = 15 ### Step 1: Write the equation for beat frequency Since l_1 < l_2$l_1 < l_2$, the frequency f_1 > f_2$f_1 > f_2$. Hence: textBeat frequency = f_1 - f_2 = fracv4l_1 - fracv4l_2$\text{Beat frequency} = f_1 - f_2 = \frac{v}{4l_1} - \frac{v}{4l_2}$ 15 = fracv4 left( frac1l_1 - frac1l_2 right)$15 = \frac{v}{4} \left( \frac{1}{l_1} - \frac{1}{l_2} \right)$ ### Step 2: Solve for velocity of sound (v$v$) Substitute the lengths in meters: 15 = fracv4 left( frac11.0 - frac11.2 right)$15 = \frac{v}{4} \left( \frac{1}{1.0} - \frac{1}{1.2} \right)$ 15 = fracv4 left( 1 - frac56 right)$15 = \frac{v}{4} \left( 1 - \frac{5}{6} \right)$ 15 = fracv4 left( frac16 right)$15 = \frac{v}{4} \left( \frac{1}{6} \right)$ 15 = fracv24$15 = \frac{v}{24}$ v = 15 times 24 = 360mathrm~m/s$v = 15 \times 24 = 360\mathrm{~m/s}$ ### Pattern Recognition Beat problems involving standing waves in organ pipes can be calculated faster by remembering that f propto frac1l$f \propto \frac{1}{l}$. This allows setting up the proportion v = 4 cdot Delta f cdot fracl_1 l_2l_2 - l_1$v = 4 \cdot \Delta f \cdot \frac{l_1 l_2}{l_2 - l_1}$ directly as a short-cut. ### 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 v = sqrtfracTmu$v = \sqrt{\frac{T}{\mu}}$ mu = rho A = rho left(pi R^2right)$\mu = \rho A = \rho \left(\pi R^2\right)$ where, v$v$ = velocity of transverse wave on a string T$T$ = tension on string mu$\mu$ = linear mass density rho$\rho$ = density of material A$A$ = cross-sectional area ### Core Logic Since both strings are made of the same material, their density rho$\rho$ is the same. Also, they are stretched to have the same amount of tension T$T$. Substitute the formula for mu$\mu$ into the velocity equation: v = sqrtfracTrho pi R^2 = frac1R sqrtfracTrho pi$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$ Thus, the wave velocity is inversely proportional to the radius of the cross section of the string: v propto frac1R$v \propto \frac{1}{R}$ fracv_2v_1 = fracR_1R_2$\frac{v_2}{v_1} = \frac{R_1}{R_2}$ ### Step 1: Ratio Computation Given: - R_1 = R$R_1 = R$ - R_2 = R/2$R_2 = R/2$ Substitute these values: fracv_2v_1 = fracRR/2 = 2$\frac{v_2}{v_1} = \frac{R}{R/2} = 2$ ### Pattern Recognition Sees: Circular cross-section strings + transverse wave speed relation. Shortcut: Wave velocity v propto frac1sqrtmu propto frac1R$v \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{R}$. If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by sqrt4 = 2$\sqrt{4} = 2$ times. ✓ ### 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