NEET Physics Question

A spring elongated by length L when a mass M is suspended to it. Now a tiny mass m is attached and then released, its time period of oscillation is: (1999)

$ 2\pi \sqrt{\frac{(M+m)\ell}{Mg}} $

$ 2\pi \sqrt{\frac{m\ell}{Mg}} $

$ 2\pi \sqrt{\frac{L}{g}} $

$ 2\pi \sqrt{\frac{M\ell}{(m+M)g}} $

Solution:

The spring constant is $ k = \frac{Mg}{L} $. When total mass becomes $ (M+m) $, the time period is $ T = 2\pi \sqrt{\frac{M+m}{k}} = 2\pi \sqrt{\frac{(M+m)L}{Mg}} $.

Rankers Physics

Solution:

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