A cylindrical block of mass M and area of cross section A is floating in a liquid of density rho$\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation 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 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 acceleration 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 angular displacement (alpha propto fracthetal$\alpha \propto \frac{\theta}{l}$). To maintain identical angular accelerations, the ratio fracthetal$\frac{\theta}{l}$ must remain constant, yielding the cross-multiplication relation theta_1 l_2 = theta_2 l_1$\theta_1 l_2 = \theta_2 l_1$. ### Evaluation Rubric / Model Answer Option D: theta_1l_2 = theta_2l_1$\theta_{1}l_{2} = \theta_{2}l_{1}$ ### Chapter Mix Class 11 Physics: Oscillations