A block of mass M is placed on top of a hole in a horizontal table. A spring of force constant k is connected to the block through the hole. The other end of the massless spring has a particle of mass m connected to it. With what maximum amplitude can the particle oscillate up and down such that the block does not lose contact with the table?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{M g}{k}$

$\frac{(M + m) g}{k}$

$\frac{(M - m) g}{k}$

$\frac{m g}{k}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The block will lose contact if the spring compresses more than $\frac{Mg}{k}$ . In equilibrium position the spring is stretched by $\frac{mg}{k}$ If the spring oscillates with amplitude A , it will move up above its equilibrium position by A. It means compression in this extreme position will be $A - \frac{mg}{k}$ For the block to remain on the table – $A - \frac{Mg}{k} = \frac{mg}{k}$
$A = \frac{Mg}{k} + \frac{mg}{k}$