A 1.50-kg ball and a 2.00-kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant 165 N/m. and the system is vibrating vertically with amplitude 15.0 cm. The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the ?lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.
Solution 39E Step 1 of 5: (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion At the lowest point the acceleration due to gravity acting on the lighter ball of mass 1.5kg will drag it downward, whereas the heavier ball of mass 2 kg which is directly connected to the spring with force constant k=165 N/m will be acted by the tension force which is opposite to the gravity and therefore the system of balls vibrates with amplitude A=15 cm=0.15 m. Step 2 of 5: (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose. Let L oe the length of the unstretched spring with force constant k=165 N/m. The equilibrium position when both balls were attached, Mass attached, M=2 kg +1.5 kg=3.5 kg force constant k=165 N/m Equilibrium position = original length + extension due to weight X =L +o Using F=W=kx as x=W/k X=L +oW/k) Substituting k=165 N/m , W=Mg=3.5(9.8) X=L +( 3.5(9.).................1 o 165 Step 3 of 5: The new equilibrium position when only heavier is attached Mass, m=2kg 1 X =L +(o/k) Substituting k=165 N/m , W=mg=2(9.8) X =L +o 2(9.)....................2 165