​​
​​
A spring with natural length (N . L .) hangs from the ceiling . A mass
m
is placed on the spring and is in equilibrium at position O (E . P . -Equilibrium Position), at a distance of 1 meter from the floor . ​The mass oscillates, starting from position O, executing oscillation between the two extreme positions A and B, which are a fixed distance apart equal to
2
y
0
meters . The distance of the mass (in meters) from the floor, as a function of time (in seconds), is given by :
y(t)1+0.2sin
π
2
t
1
.
Find
y
0
and then find the distance between the two extreme positions A and B of the oscillation .
2
.
Find the period of the oscillation .
3
.
Plot the function for
t
in the interval [0, 4] .
4
.
Find at what times the distance of the mass from the floor will be equal to 1.1 meters, for
t
in the interval [0, 2] .

​
Solution

1
.
The trigonometric function
f(x)ρ·sin(ωx)
, where
ρ,ω>0
has maximum
ρ
, minimum
-ρ
and period
T
2π
ω
y
0
(OA)(OB)0.2
meters
In[]:=
r=0.2;
The function
y(t)
has maximum
In[]:=
max=1+r
Out[]=
1.2
and minimun
In[]:=
min=1-r
Out[]=
0.8
The distance between the two extreme positions A and B is
In[]:=
Abs[max-min]
Out[]=
0.4
2
.
The period of the function
y(t)
is
In[]:=
T=2Pi/(Pi/2)
Out[]=
4
3
.
​
In[]:=
y[t_]=1+0.2Sin[Pi/2t]
Out[]=
1+0.2Sin
πt
2

In[]:=
table=Table[{t,y[t]},{t,0,4}];​​
In[]:=
TableForm[table,TableHeadings->{None,{"t","y(t)"}}]
Out[]//TableForm=
t
y(t)
0
1.
1
1.2
2
1.
3
0.8
4
1.
In[]:=
Plot[y[t],{t,0,4},PlotStyle->{Green,Thick},PlotRange->{{0,4},{0,Automatic}},PlotLabels->Automatic,AxesLabel->{"x","y"}]
Out[]=
4
.
​
Essentially, we are asking to find the time moment
t∈[0,2]
when
y(t)1.1
.
In[]:=
​​Reduce[{0<=t<=2,y[t]==1.1},t]
Reduce
:Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.
Out[]=
t0.333333||t1.66667
So
t
1
3
and
t
5
3
.