A mass of 3.25 kg is attached to the end of a spring that is stretched 22 cm by a force of 15 N. It is set in motion with an initial position x0=0 and initial velocity (in m/s) v0=−12. Find the amplitude, period, and frequency of the resulting motion. You should assume that the system is frictionless.

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,

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: