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You are watching an object that is moving in SHM. When the object is displaced 0.650 m to the right of its equilibrium position, it has a velocity of 2.50 m/s to the right and an acceleration of 8.40 m/s2 to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Respuesta :

davidlcastiblanco

Answer:

Explanation:

Let's model with sine:

x(t) = Asin(ωt)

so that t = 0, x = 0

x(t) = 0.65 m = A*sin(ω*t)

v(t) = x(t)'= 2.50 m/s = A*ω*cos(ω*t)

a(t) = v(t)'= -8.40 m/s² = -A*ω²*sin(ω*t)

x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)

0.65m / -8.40 m/s² = -1 / ω²

ω² = 12.934 rad^2/s^2

ω = 3.59 rad/s  

x(t) / v(t) = Asin(ωt) / Aωcos(ωt)

0.650m / 2.50m/s = tan(3.59t) / 3.59 rad/s

0.9334 = tan(3.59t)

t = 0.176 s  

x(0.176) = A*sin(3.59*0.176)

0.65 m= A*sin(0.631)

A = 0.732 m ← amplitude of motion

marianaegarciaperedo

The SHM the oscillating and periodic motion of an object. The distance that the object needs to move from x(t) = 0.65m to x(t) = A = 1.182m is 0.532m.

What is the simple harmonic motion?

It is a rectilinear movement performed by a oscillating and periodic movil.

When considering a simple harmonic motion, we need to know that there will be always a restoring force (F), that tends to take the movil to the original position when it moves a distance named amplitud.

There are some significant components to consider in this motion,

  • The position of the movil, X (m) ⇒ initial, equilibrium, and in other positions
  • The acceleration, a (m/s²)
  • Velocity, V (m/s)
  • The amplitud, A (m)
  • The cyclic frequency, w (rad/s)
  • The initial phase, Φ (rad)

The distance that the object needs to move from x(t) = 0.65m to x(t) = A = 1.182m is 0.532m.

NOTE: Due to technical problems, you will find the complete explanation in the attached files.

You can learn more about simple harmonic motion at

https://brainly.com/question/13858183

https://brainly.com/question/17315536

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