Respuesta :
Answer:
Step-by-step explanation:
Given that,
Mass of object=3.25kg
The extension e=22cm=0.22m
Force applied to cause extension F=15N
Initial position Xo=0
Initial velocity Vo=-12m/s
We can get the spring constant from Hooke's law
F=ke
Then, k=F/e
k=15/0.22
k=68.182N/m
Also our natural frequency w is given as
w=√(k/m)
Therefore,
w=√(68.182/3.24)
w=√20.98
w=√21
w=4.58rad/s
w=4.6rad/s
There is no damping in this situation, no outside force acting on the system and the equation that governs the system is
mx''+kx=0
3.25x''+68.182x=0
Divide through by 3.25
x''+20.98x=0
We can approximate 20.98 to 21
x"+21x=0
The solution to this differential equation using D operator
D²+21=0
D²=-21
D= ±√-21
D=±√21 •i
Then the solution is
x(t)=A•Sinwt +B•Coswt
x(t)=A•Sin√21 t +B•Cos√21 t
Note that x'(t)=v(t)
and at t=0 Vo=-12m/s
x(t)=A•Sin√21 t +B•Cos√21 t
x'(t)=v(t)=A√21•Cos√21 t - B√21•Sin√21 t
v(t)=A√21•Cos√21 t - B√21•Sin√21 t
Then, using the two initial conditions
v(0)=-12
And X(0)=0
x(t)=A•Sin√21 t +B•Cos√21 t
X(0)=A•Sin√21•0 +B•Cos√21•0
X(0)=A•Sin0+B•Cos0
0=B
B=0
Also,, V(0)=-12m/s
v(t)=A√21•Cos√21 t - B√21•Sin√21 t
V(0)=A√21•Cos√21•0- B√21•Sin√21•0
V(0)=A√21•Cos0- B√21•Sin0
-12=A√21
Therefore,
A=-12/√21
A=-2.62
Therefore the general equation becomes
x(t)=A•Sin√21 t +B•Cos√21 t
x(t)=-2.62Sin√21 t +0•Cos√21 t
x(t)=-2.62Sin√21 t
a. The amplitude
Comparing x(t) to wave equations
x(t)=-Asin(wt+2λ/t)
Then,
A=2.62m
b. We know the natural frequency already to be
w=√21
w=4.58rad/s
c. Period
Comparing the equation again
wt=√21t
Given that w=2πf
Therefore, 2πft=√21t
Then, f=√21t / 2πt
f=√21/2π
f=0.73Hz
Then, period is the reciprocal of frequency
T=1/f
T=1/0.73
T=1.37seconds
The period is 1.37sec,
Following are calculations to the given question:
Amplitude, period, and frequency calculation:
mass [tex](m)= 3.25 \ kg\\\\[/tex]
stretcher [tex](x)=22\ cm\\[/tex]
Force [tex](f)= 15\ N\\\\[/tex]
initial position [tex]x_0=0 \\\\[/tex]
initial velocity [tex]v_0=-12[/tex]
Using the formula for calculating the value:
[tex]\to F=kx =15\ N \\\\\to k = \frac{15}{0.22} =68.18\\\\[/tex]
[tex]\to A=v \times \sqrt{(\frac{m}{k})}[/tex]
[tex]=5\times \sqrt{\frac{3.25}{68.18}} \\\\= 5\times \sqrt{0.04766}\\\\= 5\times 0.2183 \\\\=1.09\ m[/tex]
[tex]\to T =2\times \pi \times \sqrt{(\frac{m}{k})}[/tex]
[tex]=2\times 3.14 \times 0.2183 \\\\=1.37 \ sec\\\\[/tex]
[tex]\to f=\frac{w}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
[tex]=\frac{1}{2\times 3.14}\sqrt{\frac{68.18}{3.25}}\\\\ =\frac{1}{2\times 3.14}\sqrt{20.97}\\\\=\frac{1}{2\times 3.14}\times 4.58\\\\= \frac{1}{3.14}\times 2.29\\\\=0.729\ Hz\\\\[/tex]
Therefore, the final answer is "[tex]\bold{1.09\ m, 1.37\ sec, \ and \ 0.729\ Hz}[/tex]".
Find out more information about the amplitude, period, and frequency here:
brainly.com/question/1687440
