A block of mass m is at rest on another block | vedclass

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(C) Let the total mass be $M = 2m$. The equilibrium position of the system is where the spring is compressed by $x_0 = \frac{Mg}{K} = \frac{2mg}{K}$.F

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$\frac{2mg}{K}$

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(C) Let the total mass be $M = 2m$. The equilibrium position of the system is where the spring is compressed by $x_0 = \frac{Mg}{K} = \frac{2mg}{K}$.For the blocks to remain in contact,the normal force $N$ between them must be $\ge 0$.The equation of motion for the upper block is $mg - N = ma$,where $a$ is the acceleration of the system.For the system,the restoring force is $F = -Kx$,where $x$ is the displacement from the equilibrium position. The acceleration is $a = -\frac{Kx}{M} = -\frac{Kx}{2m}$.Substituting $a$ into the equation for the upper block: $N = mg - m(-\frac{Kx}{2m}) = mg + \frac{Kx}{2}$.For $N \ge 0$,we need $mg + \frac{Kx}{2} \ge 0$,which is always true for $x \ge -x_0$.The critical condition occurs at the highest point of the oscillation,where the acceleration is downward and maximum. The maximum downward acceleration is $a_{max} = \omega^2 A = \frac{K}{2m} A$.The upper block remains in contact as long as its downward acceleration does not exceed $g$. Thus,$a_{max} \le g$.$\frac{K}{2m} A \le g \implies A \le \frac{2mg}{K}$.Therefore,the maximum amplitude is $A = \frac{2mg}{K}$.

$A$ block of mass $m$ is at rest on another block of the same mass as shown in the figure. The lower block is attached to a spring. What is the maximum amplitude of motion so that both blocks remain in contact?

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