A 220 g block hangs from a spring with spring constant16 N/m. At t=0 the block i
ID: 1727963 • Letter: A
Question
A 220 g block hangs from a spring with spring constant16 N/m. At t=0 the block is 18 cm below the equilibriumpoint and moving upward with a speed of 122 cm/s. What are theblock's....a) oscillation frequency?
b) distance from equilibrium when the speed is72 cm/s?
c) position at t=5 s? A 220 g block hangs from a spring with spring constant16 N/m. At t=0 the block is 18 cm below the equilibriumpoint and moving upward with a speed of 122 cm/s. What are theblock's....
a) oscillation frequency?
b) distance from equilibrium when the speed is72 cm/s?
c) position at t=5 s?
Explanation / Answer
angularfrequency = (k/m) = (16/0.220) = 8.258 rad/s a. Frequency f = /2 = 8.258/ 2 *3.14 = 1.31 Hz b. speed v2 = 2* ( A2 - y2 ) at y = 18cm v18 = 1222 = 8.2582*(A2 - 182) A2 = (1222/8.2582) + 182 = 542.26 => amplitude A = 542.26 = 23.28 cm at v = 72 cm 722 = 8.2582* (542.26 - y2) y = 21.59 cm c. position y = A* sin * t = 23.28 * sin(8.258 *5) = -10.11 cmRelated Questions
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