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(B) Let the tension in the string be $T$. From the force balance on the block,the tension $T = mg$.For the pulleys and springs,let the extensions in t

Vedclass · Accepted Answer

The angular frequency for small oscillation of the system is $\sqrt{\frac{4K}{33m}}$.

Vedclass · Answer

(B) Let the tension in the string be $T$. From the force balance on the block,the tension $T = mg$.For the pulleys and springs,let the extensions in the three springs be $x_1, x_2, x_3$ respectively. The effective spring constant $K_{eff}$ is determined by the relation $x = F/K_{eff}$,where $x$ is the displacement of the block.Using the constraint relation for the pulleys,the displacement of the block $x$ is related to the extensions as $x = \frac{x_1}{2} + 2x_2 + 2x_3$.Given the tension $T$ in the string,the forces on the springs are $F_1 = T/2$,$F_2 = 2T$,and $F_3 = 2T$.Thus,$x_1 = \frac{T}{2K}$,$x_2 = \frac{2T}{K}$,and $x_3 = \frac{2T}{K}$.Substituting these into the constraint equation: $x = \frac{T}{4K} + \frac{4T}{K} + \frac{4T}{K} = \frac{T + 16T + 16T}{4K} = \frac{33T}{4K}$.Since $T = mg$,$x = \frac{33mg}{4K}$. The effective spring constant is $K_{eff} = \frac{mg}{x} = \frac{4K}{33}$.The angular frequency $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{4K}{33m}}$.The elastic potential energy at equilibrium is $U = \frac{1}{2} K_1 x_1^2 + \frac{1}{2} K_2 x_2^2 + \frac{1}{2} K_3 x_3^2 = \frac{1}{2} K (\frac{mg}{2K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 + \frac{1}{2} K (\frac{2mg}{K})^2 = \frac{m^2g^2}{8K} + \frac{2m^2g^2}{K} + \frac{2m^2g^2}{K} = \frac{m^2g^2 + 16m^2g^2 + 16m^2g^2}{8K} = \frac{33m^2g^2}{8K}$.

Assuming all pulleys,springs,and strings are massless. Consider all surfaces smooth. Choose the correct statement$(s)$.

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