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Q. An ideal spring with spring-constant $K$ is hung from the ceiling and a block of mass $M$ is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

Oscillations

Solution:

Let x be the maximum extension of the spring. From energy conservation
image
Loss in gravitational potential energy
$=$ Gain in potential energy of spring
$M g x=\frac{1}{2} K x^{2}$
$\Rightarrow \quad x=\frac{2 M g}{K}$