a 1.50-kg ball and a 2.00-kg ball are glued together with the lighter one below

Question

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 the lowest position in their motion.
(a) Why is the glue more likely to fail at the lowest point that at any other point in motion?
(b)FInd the amplitude and frequency of the vibrations after the lower ball has come loose. <- I can't find amplitude??

Explanation / Answer

Let Lo be the length of the un-stretched spring The equilibrium position when both balls were attached was Lo+(3.5*9.8)/165 m from the support. The new equilibrium position when only heavier is attached = Lo+(2*9.8)/165 m So the equilibrium position got shifted up by [(1.5*9.8)/165] m = 0.0875. Hence new amplitude = 0.15 + 0.0875 m = 0.2375 or 23.8 cm The new frequency, f is given by f = [1/(2*pi)]*sq rt(165/2.0) = 1.446 or 1.45 Hz

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