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denpristay [2]
3 years ago
15

An object is in simple harmonic motion. Find an equation for the motion given that the frequency is 3⁄π and at time t = 0, y = 0

, and y' = 6. What is the equation of motion?
Physics
2 answers:
alexdok [17]3 years ago
7 0

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation:

because im smrt

Delicious77 [7]3 years ago
3 0

Answer: y(t)= 1/π^2 sin(6*π^2*t)

Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:

y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.

The data are: ν= 3π; y(t=0)=0 and y'(0)=6.

Firstly we know that 2πν=ω then ω=6*π^2

Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0

Besides y'(t)=6*π^2*A*cos (6*π^2*t)

y'(0)=6=6*π^2*A*cos (6*π^2*0)

6=6*π^2*A then A= 1/π^2

Finally the equation is:

y(t)= 1/π^2 sin(6*π^2*t)

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To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a ho
Fudgin [204]

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the coefficient of friction is 0.32

Explanation:

 Given the data in the question;

we make use of kinematic equation of motion;

ω = ω₀ + ∝t

we substitute

ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )

ω = 3.9612 rad/s

The centripetal force acting on the sample is;

Fc = mrω²

from the image; r = 200 mm = 0.2 m

so we substitute

Fc = m(0.2 m ) ( 3.9612 rad/s )²

Fc = (3.13822 m/s²)m

we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;

f = Fc

μN = Fc

μmg = (3.13822 m/s²)m

μ = (3.13822 m/s²)m / mg

μ = (3.13822 m/s²) / g

acceleration due to gravity g = 9.8 m/s²

so

μ = (3.13822 m/s²) / 9.8 m/s²

μ = 0.32

Therefore, the coefficient of friction is 0.32

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3 years ago
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