Rankers Physics

Spring Block System: Practice Problem & Solution

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 Explained:

To solve this problem, we apply the core principles of Spring Block System. Understanding the underlying formula is key to arriving at the correct answer below:

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}} $.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): For a physical pendulum if distance of point of suspension from centre of mass increases time period first decreases then increases. Reason (R): For a physical pendulum there is some distance from centre of mass at which frequency of oscillation is maximum.  
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