Can some one do my homework plz

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 ini

mylen [45]

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,

5 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Rewrite the fol

never [62]

Vertex Form: y = a(x - h)^2 + k, with (h,k) as the vertex.

To convert it into vertex form, we are going to be completing the square. Firstly, subtract 56 on both sides: $y-56=2x^2-32x$

Next, factor out the GCF, or greatest common factor, of the right side. To find the GCF, list the factors of both terms and the greatest one they share is their GCF. In this case, the GCF is 2: $y-56=2(x^2-16x)$

Next, we want to make what's inside the parentheses a perfect square. To find the constant of the perfect square, divide the x term by 2 and square the quotient. In this case:

-16 ÷ 2 = -8, (-8)² = 64

So you're gonna have to add 64 to the inside of the parentheses. To cancel this out, you will need to add to the other side the product of 64 and 2 (since 64 is inside the parentheses), which is 128: $y+72=2(x^2-16x+64)$