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January 27, 2015

January 27, 2015

Posted by **anon** on Friday, October 11, 2013 at 4:40am.

(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 2:32pmω₀=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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