### Related Formula For combined system performing simple harmonic motion without relative slipping: T = 2pi sqrtfracm_texttotalk$T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}}$ a = -omega^2 x = -frackM+m x$a = -\omega^2 x = -\frac{k}{M+m} x$ ### Core Logic Let's analyze each statement: - **Statement (A)**: Since both blocks perform SHM together, the combined mass is (M + m)$(M + m)$. The spring constant is k$k$. Thus, the time period of small oscillation is: T = 2pi sqrtfracM+mk$T = 2\pi \sqrt{\frac{M+m}{k}}$ This is correct. (A is True) - **Statement (B)**: When the system is displaced by x$x$, the restoring spring force on the combined system is F = -kx$F = -kx$. The common acceleration of the combined mass is: a = -frackxM+m implies |a| = frack|x|M+m$a = -\frac{kx}{M+m} \implies |a| = \frac{k|x|}{M+m}$ This matches the expression (taking magnitude). (B is True) - **Statement (C)**: The upper block of mass m$m$ moves solely due to the static frictional force f$f$ acting on it. Thus: f = m a = m left( frackxM+m right) = fracmkxM+m$f = m a = m \left( \frac{kx}{M+m} \right) = \frac{mkx}{M+m}$ Statement (C) claims the frictional force is fracmmu|x|M+m$\frac{m\mu|x|}{M+m}$, which is incorrect because friction is determined by acceleration, not by coefficient of friction mu$\mu$ during static grip. (C is False) - **Statement (D)**: For no slipping to occur, the maximum frictional force required at peak amplitude A$A$ must be less than or equal to the limiting static friction f_L = mu mg$f_L = \mu mg$: f_textmax = fracmkAM+m le mu mg$f_{\text{max}} = \frac{mkA}{M+m} \le \mu mg$ frackAM+m le mu g implies A le fracmu(M+m)gk$\frac{kA}{M+m} \le \mu g \implies A \le \frac{\mu(M+m)g}{k}$ Thus, the maximum amplitude is fracmu(M+m)gk$\frac{\mu(M+m)g}{k}$. (D is True) - **Statement (E)**: The maximum static frictional force between the blocks is f_L = mu mg$f_L = \mu mg$, not mu (M+m)g$\mu (M+m)g$. (E is False) ### Step 1: Conclusion Only statements A, B, and D are correct. Hence, the correct option is (1).

### Pattern Recognition In stacked blocks with springs, always identify the force driving the non-spring-loaded block. Here, mass m$m$ is driven purely by friction, so f = m cdot a$f = m \cdot a$. Slipping begins when this required force exceeds f_textlimit = mu m g$f_{\text{limit}} = \mu m g$. This simple boundary matches the derivation of maximum amplitude perfectly! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion: Friction Class 11 Physics: Oscillations: Simple Harmonic Motion