A 0.150 kg block is suspended from a spring. When a small pebble of mass 34 g is

Question

A 0.150 kg block is suspended from a spring. When a small pebble of mass 34 g is placed on the block, the spring stretches an additional 6.0 cm. With the pebble on the block, the block oscillates with an amplitude of 12.0 cm.  (Assume that the pebble is glued to the block.)

(a) What is the frequency of the motion?

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

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


This question is really frustating me.. I made all the conversions, ie. g to kg, cm to m.. and I understand I need to use y=mg/k to find k.  and then plug into omega=sqrt(k/m) to be able to solve frequency.. but.. I am doing something wrong..

PLEASE Explain.. full points for best answer

Explanation / Answer

First find out the spring constant k. It can be found from extension caused by the additional weight of the pebble

k * y = 0.034 * 9.8 => k * 0.06 = 0.3332 => k = 5.553N/m

The combined mass of the block + pebble = 150 + 34 = 184 g = 0.184 kg

a) Time period = 2 * pi * sqrt(m/k) = 1.1437 s. So, frequency = 1/T = 0.8743Hz

Angular frequency = 2 * pi / T = 5.493

b) Lowest to highest point would be covered in half of the time period = 1.1437/2 = 0.57185

c) At its highest point, the net acceleration of the block + pebble = 0.12 * 5.553 / 0.184 = 3.621 m/s^2

So, net force on the pebble = 0.034 * 3.621 = 0.1231m/s^2

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