A 0.110 kg block is suspended from a spring. When a small pebble of mass 32 g is

Question

A 0.110 kg block is suspended from a spring. When a small pebble of mass 32 g is placed on the block, the spring stretches an additional 4.8 cm. With the pebble on the block, the block oscillates with an amplitude of 11.0 cm. (Assume that the pebble is glued to the block.)
(a) What is the frequency of the motion?
Hz

(b) How long does the block take to travel from its lowest point to its highest point?
s

(c) What is the net force on the pebble when it is at a point of maximum upward displacement?
N (downward)

Explanation / Answer

a) frequency of the system             f= 1/2 * mg/y mtotal                = 1/2 ( 3.14 ) * 0.032 kg * 9.8 m/s2/0.048m *0.142 kg                = 12.57/6.28                f= 1.080 Hz      b)   t = T/2               = 1/2f              = 0.46 s c) fnet = 42 f2 mA               = 4 ( 3.14)2 ( 1.080 Hz)2 * 0.032 kg * 0.11m               = 0.1619 N                = 1/2 ( 3.14 ) * 0.032 kg * 9.8 m/s2/0.048m *0.142 kg                = 12.57/6.28                f= 1.080 Hz      b)   t = T/2               = 1/2f              = 0.46 s c) fnet = 42 f2 mA               = 4 ( 3.14)2 ( 1.080 Hz)2 * 0.032 kg * 0.11m               = 0.1619 N

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