A 0.100 kg block is suspended from a spring. When a small pebble of mass 40 g is

Question

A 0.100 kg block is suspended from a spring. When a small pebble of mass 40 g is placed on the block, the spring stretches an additional 5.0 cm. With the pebble on the block, the block oscillates with an amplitude of 8.0 cm. (Assume that the pebble is glued to the block.)
(a) What is the frequency of the motion?
in Hz

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

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

Explanation / Answer

Mass of the block = 0.100 Kg

Mass of the pebble= 0.040 Kg

Extension of spring= 0.05

as the spring force equals the weight

k*(0.05)=0.040 *10
k=8 N/m

a) The angular frequency is = (k/m) = 8/0.14 = 7.56 Hz

b)Time period= = 0.83 s

It requires one-fourth of the time to go grom lower position to higher position

so time required is = 0.20 sec

c)Amplitude(A)= 0.08m

The net force at maximum upward position is mw2A+mg= 0.140 * 8* 0.08 /0.14 + 1.4= 2.04N

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