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?

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.

To answer these questions, we need to consider the equations of motion for the spring-mass system.

1) To find the position of the block as a function of time, we can use the equation of motion for simple harmonic motion:

x(t) = A * cos(ωt + φ),

where x(t) is the position of the block at time t, A is the amplitude of the motion, ω is the angular frequency (ω = √(k/m)), and φ is the phase constant.

In this case, the block is initially compressed a distance of x0 = 1.7 m from equilibrium, so the amplitude is A = x0. The angular frequency ω is given by ω = √(k/m), where k is the spring constant (k = 20 N/m) and m is the mass of the block (m = 6 kg).

To find the phase constant φ, we need to consider the initial conditions of the system. The block is released with an initial velocity v0 = 5 m/s towards the equilibrium position. At t = 0, the displacement x(0) = -x0 (since the block is compressed), and the velocity v(0) = v0.

Using these initial conditions, we can write the equations:

x(0) = A * cos(φ) = -x0,
v(0) = -A * ω * sin(φ) = v0.

Solving these equations simultaneously, we can find the phase constant φ. Once we have φ, we can express the position of the block as a function of time.

2) To find the time it takes for the mass to first return to the equilibrium position, we need to find the period of the motion. The period T is defined as the time it takes for the system to complete one full oscillation. It is given by:

T = 2π/ω.

3) To find the time it takes for the spring to first become completely extended, we need to consider the point when the block reaches its maximum displacement from equilibrium. At this point, the block momentarily comes to rest with its velocity being zero. We can use the equation of motion:

v(t) = -A * ω * sin(ωt + φ).

Setting v(t) = 0, we can solve for the time when the velocity becomes zero to find the time it takes for the spring to become completely extended.

By following these steps, we can answer the given questions about the spring-mass system.