A block of mass $m$ is suspended from the ceiling of a stationary standing elevator through a spring of spring constant $k$ . Suddenly the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion in the elevator. Find the amplitude.

\frac{m g}{k}

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\frac{2 m g}{3 k}

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\frac{k}{m g}

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\frac{m g}{2 k}

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Solution

The correct option is A $\frac{m g}{k}$
When elevator is stationary spring is stretched to support block weight, if elongation is A then, $m g = k A$ .
When cable breaks, the elevator starts falling with acceleration $g$ and the elevator becomes an accelerated (non inertial) frame of reference. In that frame pseudoforce comes into play. This pseudoforce balances the weight and a net force $k x$ acts on the system where the acceleration $\frac{k x}{m}$ is proportional and oppositely directed to the displacement.
Hence the motion is simple harmonic.
In initial case, the spring was stretched by A where velocity was zero. This elongation should give us the amplitude as well.
Hence, $A = \frac{m g}{k}$

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A block of mass m is suspended from the ceiling of a stationary standing elevator through a spring of spring constant k. Suddenly the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion in the elevator. Find the amplitude.

A block of mass $m$ is suspended from the ceiling of a stationary standing elevator through a spring of spring constant $k$ . Suddenly the cable breaks and the elevator starts falling freely. Show that the block now executes a simple harmonic motion in the elevator. Find the amplitude.