Posted by **PLS HELP** on Friday, October 11, 2013 at 9:46am.

Consider an ideal spring that has an unstretched length l0 = 3.9 m. Assume the spring has a constant k = 20 N/m. Suppose the spring is attached to a mass m = 6 kg that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x0 = 1.7 m from equilibrium and then released with an initial speed v0 = 5 m/s toward the equilibrium position.

(1)What is the position of the block as a function of time. Express your answer in terms of t.

(2)How long will it take for the mass to first return to the equilibrium position?

(3)How long will it take for the spring to first become completely extended?Consider an ideal spring that has an unstretched length l0 = 3.9 m. Assume the spring has a constant k = 20 N/m. Suppose the spring is attached to a mass m = 6 kg that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x0 = 1.7 m from equilibrium and then released with an initial speed v0 = 5 m/s toward the equilibrium position.

(1)What is the position of the block as a function of time. Express your answer in terms of t.

(2)How long will it take for the mass to first return to the equilibrium position?

(3)How long will it take for the spring to first become completely extended?

- physics -
**Elena**, Friday, October 11, 2013 at 2:31pm
ω₀=sqrt(k/m) = sqrt(20/6) =1.83 rad/s

T=2π/ ω =2π/1.83 =3.43 s.

(1)

x=Acos (ω₀t +α)

At t=0

-x₀=x

v₀=v(x)

v(x)= dx/dt = - ω₀Asin(ω₀t +α) … (1)

If t=0

-x₀=Acos (ω₀t +α) =Acosα…..(2)

and

v(x)= - ω₀Asin(ω₀t +α) =- ω₀Asin α…(3)

From (3)

Asin α =- v₀/ω₀ ……(4)

Divide (4) by (2)

Asin α/ Acosα = v₀/x₀•ω₀ =>

tan α= v₀/x₀•ω₀ = 5/1.7•1.83 =1.607

α=58° ≈1 rad

The square of (4) + the square of (2)

(Asin α)² +(Acosα)² =A²= (x₀)²+(v₀/ω₀)² =>

A=sqrt[(x₀)²+(v₀/ω₀)²] =

=sqrt{1.7² +(5/1.83)²}=3.22 m

The position of the object-spring system is given by

x(t) =3.22cos(1.83t+1) (m)

(2)

The spring first reaches equilibrium

at time t₁

x(t₁) = 0 =>

x(t₁) =3.22cos(1.83t₁+1) = 0

cos(1.83t₁+1)=0

1.83t₁+1 = π/2

t₁=[(π/2) -1)]/1.83=0.31 s.

(3)

The object is first completely extended when the velocity is zero.

v(x)= - ω₀Asin(ω₀t +α)=

=1.83•3.22sin(1.83t₂+1) =0,

sin(1.83t₂+1)=0

1.83t₂+1= π

t₂=(π-1)/1.83 =1.17 s.

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