### Related Formula v_max = Aomega$v_{\max} = A\omega$ omega = sqrtfrackm$\omega = \sqrt{\frac{k}{m}}$ ### Core Logic The maximum velocity in vertical SHM happens at the mean equilibrium position and is given by v_max = Aomega$v_{\max} = A\omega$. Given m_A = m_B = m$m_A = m_B = m$ and A_A = A_B = A$A_A = A_B = A$: v_A = A omega_1 = A sqrtfrack_1m$v_A = A \omega_1 = A \sqrt{\frac{k_1}{m}}$ v_B = A omega_2 = A sqrtfrack_2m$v_B = A \omega_2 = A \sqrt{\frac{k_2}{m}}$ Taking the ratio: fracv_Av_B = fracomega_1omega_2 = fracsqrtfrack_1msqrtfrack_2m = sqrtfrack_1k_2$\frac{v_A}{v_B} = \frac{\omega_1}{\omega_2} = \frac{\sqrt{\frac{k_1}{m}}}{\sqrt{\frac{k_2}{m}}} = \sqrt{\frac{k_1}{k_2}}$ ### Pattern Recognition Since maximum velocity is directly proportional to angular frequency omega$\omega$ for equal amplitudes, and omega propto sqrtk$\omega \propto \sqrt{k}$, the velocity ratio directly yields the square root of the spring constant ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula Angular acceleration definition for a simple pendulum swinging at small angle theta$\theta$: alpha = -omega^2 theta$\alpha = -\omega^2 \theta$ where: omega = sqrtfracgl implies omega^2 = fracgl $\omega = \sqrt{\frac{g}{l}} implies \omega^2 = \frac{g}{l} $ Hence, the magnitude of angular acceleration is: alpha = fracgltheta$ alpha = \frac{g}{l}\theta$ ### Core Logic Given that angular accelerations are exactly identical in magnitude (alpha_1 = alpha_2$\alpha_1 = \alpha_2$): fracgl_1theta_1 = fracgl_2theta_2$\frac{g}{l_1}\theta_1 = \frac{g}{l_2}\theta_2$ ### Step 1: Simplify Expression Cancelling out the constant gravitational field factor g$g$: fractheta_1l_1 = fractheta_2l_2 implies theta_1 l_2 = theta_2 l_1$\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2} implies \theta_1 l_2 = \theta_2 l_1$ ### Pattern Recognition Angular acceleration scales inversely with length for a fixed angle. To preserve equal acceleration values, the product theta cdot l^-1$\theta \cdot l^{-1}$ must be constant, resulting in cross-multiplication balance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula x(t) = x_0 sin^2left(fract2 ight) = x_0 left(frac1 - cos t2 ight)$x(t) = x_0 \sin^2\left(\frac{t}{2} ight) = x_0 \left(\frac{1 - \cos t}{2} ight)$ ### Core Logic Given x_0 = 1\ mathrmm$x_0 = 1\ \mathrm{m}$, the position simplifies to: x = frac1 - cos t2 implies 2x - 1 = -cos t implies cos t = 1 - 2x$x = \frac{1 - \cos t}{2} \implies 2x - 1 = -\cos t \implies \cos t = 1 - 2x$ Differentiating x(t)$x(t)$ to find velocity v$v$: v = fracdxdt = frac12sin t$v = \frac{dx}{dt} = \frac{1}{2}\sin t$ Kinetic energy K$K$ is proportional to v^2$v^2$: K = frac12mv^2 = frac12m left(frac14sin^2 t ight) = fracm8(1 - cos^2 t)$K = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{1}{4}\sin^2 t ight) = \frac{m}{8}(1 - \cos^2 t)$ Substituting cos t = 1 - 2x$\cos t = 1 - 2x$: K = fracm8left[1 - (1 - 2x)^2 ight] = fracm8left[1 - (1 - 4x + 4x^2) ight] = fracm8(4x - 4x^2) = fracm2(x - x^2)$K = \frac{m}{8}\left[1 - (1 - 2x)^2 ight] = \frac{m}{8}\left[1 - (1 - 4x + 4x^2) ight] = \frac{m}{8}(4x - 4x^2) = \frac{m}{2}(x - x^2)$ This is an inverted parabola (K propto x - x^2$K \propto x - x^2$) passing through x=0$x=0$ and x=1$x=1$, reaching a peak at x = 1/2$x = 1/2$. This matches Graph (1). ### Pattern Recognition The position oscillates purely within the boundary interval [0, 1]$[0, 1]$. Kinetic energy peaks perfectly at the equilibrium midpoint x = 1/2$x = 1/2$ where speed is maximum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula In Simple Harmonic Motion: * In one full oscillation period T$T$, total distance = 4A$4A$, and net displacement = 0$0$. * In a quarter period fracT4$\frac{T}{4}$ starting from equilibrium, distance = A$A$, and displacement = A$A$. ### Core Logic Find the total number of cycles completed in the given time : textCycles = fractT = frac12.5text s2text s = 6.25text cycles = 6 text full cycles + 0.25text cycle$\text{Cycles} = \frac{t}{T} = \frac{12.5\text{ s}}{2\text{ s}} = 6.25\text{ cycles} = 6 \text{ full cycles} + 0.25\text{ cycle}$ ### Step 1: Evaluation of Distance and Displacement Calculate the total distance D$D$ covered : D = 6 times (4A) + 1A = 24A + A = 25A$D = 6 \times (4A) + 1A = 24A + A = 25A$ Given amplitude A = 1text cm$A = 1\text{ cm}$ , we get D = 25text cm$D = 25\text{ cm}$ . Calculate the net displacement d$d$: The first 6$6$ complete cycles return the particle to equilibrium with zero displacement. The remaining 0.25$0.25$ cycle (which is a quarter period fracT4$\frac{T}{4}$) moves it from the center to its maximum amplitude value : d = 1A = 1text cm$d = 1A = 1\text{ cm}$ Taking their ratio: fracDd = frac251 = 25$\frac{D}{d} = \frac{25}{1} = 25$ ### Pattern Recognition Fractional cycles modulate ratios abruptly. Always break the path index down into integer full cycles plus the final residual interval. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula The general solution tracking position in SHM is given by : x(t) = A sin(omega t + phi)$x(t) = A \sin(\omega t + \phi)$ Momentum formula: p(t) = m v(t) = m A omega cos(omega t + phi)$p(t) = m v(t) = m A \omega \cos(\omega t + \phi)$ ### Core Logic At t = 0$t = 0$ : 1) x_0 = A sin phi$x_0 = A \sin \phi$ 2) p_0 = m A omega cos phi$p_0 = m A \omega \cos \phi$ Dividing equation (1) by (2) reveals the phase \angle relationship : tan phi = left(fracx_0p_0right) m omega implies phi = tan^-1left(fracm omega x_0p_0right)$\tan \phi = \left(\frac{x_0}{p_0}\right) m \omega \implies \phi = \tan^{-1}\left(\frac{m \omega x_0}{p_0}\right)$ Squaring and combining both equations isolates amplitude A$A$: A = fracsqrt(m omega x_0)^2 + p_0^2m omega$A = \frac{\sqrt{(m \omega x_0)^2 + p_0^2}}{m \omega}$ Since both amplitude A$A$ and phase constant phi$\phi$ are explicitly fixed by the initial conditions x_0$x_0$ and p_0$p_0$, the kinematic layout of the state space is completely specified for any future time parameter t$t$. This validates that Reason (R) perfectly explains Assertion (A). ### Pattern Recognition SHM is a second-order differential equation. Any system of second-order equations requires exactly two independent boundary conditions (e.g., initial position and velocity/momentum) to completely specify unique path tracks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations