Question

A small block of mass 4.2 kg sits on top of a block of mass 19.8 kg. The lower block is attached to a spring with spring constant 248 N/m and can slide on a horizontal frictionless surface. The coefficient of friction between the blocks is 0.4.

What is the maximum possible amplitude of simple harmonic motion, x_m, of the spring-blocks system if no slippage is to occur between the blocks?

Answer 1

Ans: The situation described in problem is shown below don_ figuse m mm on M Accodding to data giyem- M = 19.8 kg M = 4.2 kg K= 248 N/m Coefficient of psichon between blocks H=:4. on this are- Now Let us focus ous attention on the small block. Fosces acting block 10 Mg Vestically downward 3) N vestically upwards due to block M 3) kN - Limiting friction at X = Xos towards horizontale We know that when a body undergoing is displaced from its equilibrium position, the force F on this body is P1 = mw²x where x- is the displacement from equilibsium position. So for the block mat Xaxom F: mw?xm Now There one force along halizontal and that is MN SHM is only - S. AN = mw? xom

N = mg continued Ako [ As there is no mation of m along vestical disection] S. ung = mw²xm Mg = w²xm Now Under the condition of no slipping Both blocks oscillate_togethes_so N W=k Putting the value of w is expression @ K M+m MTotal ug Xm к M+m Xom = MgCM+m) / K Putting values an above expression = Xm= •4x9.8 | 19.8+4.3). 248 Xoms •4x9.8 x 24 m = . .3794m 248 37.94 cm Xem = 349 m = Hence- Maximum Amplitude Xom = 37.94 com Ans
Problem is solved using basic formula of angular frequency in SHM and formula of limiting friction.