The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A sin omega t$x(t) = A \sin \omega t$ , where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T/2beta$t = T/2\beta$ . The value of beta$\beta$ is ____.

### Related Formula omega_k = 2pi f_k$$\omega_k = 2\pi f_k$$ f_textoscillator = fracf_k2$$f_{\text{oscillator}} = \frac{f_k}{2}$$ ### Core Logic For a simple harmonic oscillator with an angular frequency omega$\omega$, the kinetic energy oscillates at double the angular frequency (2omega$2\omega$). The given angular frequency of kinetic energy oscillation is omega_k = 176 text rad/s$\omega_k = 176 \text{ rad/s}$. ### Step 1: Finding Kinetic Energy Frequency First, find the frequency of kinetic energy oscillations (f_k$f_k$): f_k = fracomega_k2pi = frac1762 times frac227$$f_k = \frac{\omega_k}{2\pi} = \frac{176}{2 \times \frac{22}{7}}$$ f_k = frac176 times 744 = 4 times 7 = 28 text Hz$$f_k = \frac{176 \times 7}{44} = 4 \times 7 = 28 \text{ Hz}$$ ### Step 2: Final Conclusion Since kinetic energy oscillates at twice the frequency of the oscillator itself: f_textoscillator = fracf_k2 = frac282 = 14 text Hz$$f_{\text{oscillator}} = \frac{f_k}{2} = \frac{28}{2} = 14 \text{ Hz}$$ ### Pattern Recognition KE and PE always oscillate at twice the frequency (2nu$2\nu$) of the underlying SHM (v$v$) due to the sin^2(omega t)$\sin^2(\omega t)$ or cos^2(omega t)$\cos^2(\omega t)$ terms resolving to (1 - cos(2omega t))/2$(1 - \cos(2\omega t))/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula T = 2pi sqrtfraclg$$T = 2\pi \sqrt{\frac{l}{g}}$$ T propto sqrtl$$T \propto \sqrt{l}$$ ### Core Logic For a simple pendulum, the period is proportional to the square root of the length. If the frequency is doubled, the period is halved, which requires the length to become one-fourth of its original value. ### Step 1: Find Old and New Period Initial time period T_1 = frac1020 = 0.5 text s$T_{1} = \frac{10}{20} = 0.5 \text{ s}$ New time period T_2 = frac1040 = 0.25 text s$T_{2} = \frac{10}{40} = 0.25 \text{ s}$ So, T_2 = fracT_12$T_{2} = \frac{T_{1}}{2}$ ### Step 2: Calculate New Length fracT_1T_2 = sqrtfracl_1l_2$$\frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}}$$ 2 = sqrtfrac30l_2$$2 = \sqrt{\frac{30}{l_{2}}}$$ 4 = frac30l_2$$4 = \frac{30}{l_{2}}$$ l_2 = frac304 = 7.5 text cm$$l_{2} = \frac{30}{4} = 7.5 \text{ cm}$$ ### Pattern Recognition Sees: "Oscillations doubled in same time" → Frequency is doubled → Time period is halved. Since T propto sqrtl$T \propto \sqrt{l}$, l$l$ must be reduced to frac14$\frac{1}{4}$ of the original. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula T = 2pi sqrtfracMk_texteffective$$T = 2\pi \sqrt{\frac{M}{k_{\text{effective}}}}$$ F_textbuoyancy = rho V g$$F_{\text{buoyancy}} = \rho V g$$ ### Core Logic

At equilibrium, the buoyant force balances the weight: rho Ahg = Mg$\rho Ahg = Mg$ After displacing the block downward by a small distance x$x$, the net restoring force is: Ma = -rho A(h+x)g + Mg$$Ma = -\rho A(h+x)g + Mg$$ Ma = -rho Ahg - rho Axg + Mg$$Ma = -\rho Ahg - \rho Axg + Mg$$ Since rho Ahg = Mg$\rho Ahg = Mg$, this simplifies to: Ma = -rho Axg$Ma = -\rho Axg$ a = left(frac-rho AgMright) x$$a = \left(\frac{-\rho Ag}{M}\right) x$$ ### Step 1: Compare with Standard SHM Equation Comparing with a = -omega^2 x$a = -\omega^2 x$, we get: omega = sqrtfracrho AgM$$\omega = \sqrt{\frac{\rho Ag}{M}}$$ The time period T$T$ is: T = frac2piomega = 2pi sqrtfracMrho Ag$$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{\rho Ag}}$$ ### Pattern Recognition For a floating body of uniform cross-section A$A$, the restoring force constant is simply k = rho A g$k = \rho A g$. The time period is immediately T = 2pi sqrtm/k$T = 2\pi \sqrt{m/k}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations Class 11 Physics: Mechanical Properties of Fluids

### Related Formula omega = sqrtfrackmu$$\omega = \sqrt{\frac{k}{\mu}}$$ where mu$\mu$ is the reduced mass of the system given by: mu = fracm_1 m_2m_1 + m_2$$\mu = \frac{m_1 m_2}{m_1 + m_2}$$ ### Step 1: Calculate Reduced Mass Given m_1 = 1 text kg$m_1 = 1 \text{ kg}$ and m_2 = 0.2 text kg$m_2 = 0.2 \text{ kg}$. mu = frac1 times 0.21 + 0.2 = frac0.21.2$$\mu = \frac{1 \times 0.2}{1 + 0.2} = \frac{0.2}{1.2}$$ mu = frac16 text kg$$\mu = \frac{1}{6} \text{ kg}$$ ### Step 2: Calculate Angular Frequency Given spring constant k = 150 text N/m$k = 150 \text{ N/m}$. omega = sqrtfrac1501/6 = sqrt150 times 6 = sqrt900$$\omega = \sqrt{\frac{150}{1/6}} = \sqrt{150 \times 6} = \sqrt{900}$$ omega = 30 text rad/s$$\omega = 30 \text{ rad/s}$$ ### Pattern Recognition For any two-body spring system free of external forces, always substitute the individual mass m$m$ with the reduced mass mu = fracm_1 m_2m_1 + m_2$\mu = \frac{m_1 m_2}{m_1 + m_2}$ in the standard oscillator formula omega = sqrtk/m$\omega = \sqrt{k/m}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations