When the system shown in the diagram is in equilibrium, the right spring is stretched by 1 cm. The coefficient of static friction between the blocks is 0.3. There is no friction between the bottom block and the supporting surface. The force constants of the springs are 150 N m⁻¹ and 450 N m⁻¹ (refer figure). The blocks have equal mass of 2 kg each. Find the maximum amplitude (in cm) of the oscillations of the system shown in the figure that does not allow the top block to slide on the bottom.

Question

G1 Physics · Accepted Answer

At equilibrium the right spring is stretched 1 cm, so its force is 150×0.01=1.5 N; static friction balances this on the upper block. When both blocks move together, total stiffness=450+150=600 N m⁻¹ and total mass=4 kg, so ω²=150 s⁻². The oscillatory friction required by the top block changes by (450-2ω²)x=(450-300)x=150x. At the limiting extreme, 1.5+150A=μmg=0.3×2×10=6. Thus A=0.03 m=3 cm.