If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. suppose it is displaced 0.125 m from its equilibrium position and released with zero initial speed. after 0.860 s, its displacement is found to be 0.125 m on the opposite side and it has passed the equilibrium position once during this interval.

find the amplitude of the motion.

=

find the period of the motion.

=

find the frequency of the motion.

=

a)A=0.125 m

b)T = 1.72 s

c)f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

answer; ///i do believe that the correct answer is ; ///(physical change);

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz