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A small boy attached to one end of a vertically hanging spring is performing SHM about its mean position with angular frequency $ω$ and amplitude a. If at height y from the mean position the body gets detached from the spring. Calculate the value of y so that height obtained by the mass is maximum. The body does not interact with the spring during its subsequent motion after detachment

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Solution

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The total energy of the body will be due to gravitational potential energy, kinetic energy and elastic potential of spring.
The mass will reach the highest point when its mechanical energy [gravitational potential energy and kinetic energy] is maximum. This is possible when elastic potential energy of system is zero.
This means that the mass should detach when the spring is at its natural length.
Let L = Natural length of spring when mass m is hanging at the equilibrium.
Now, $m g = K l l = \frac{m g}{K}$
So, $y = \frac{m g}{K}$
But spring constant $K = m ω^{2}$
$y = \frac{m g}{m ω^{2}} = \frac{g}{ω^{2}}$
where $\frac{g}{ω} < a$

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