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 motion of the spring-mass system using the equations of motion. Let's break it down step by step.
1. What is the position of the block as a function of time?
The position of the block as a function of time can be determined using the equation of 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 (initial displacement from equilibrium),
- ω is the angular frequency of the motion,
- φ is the phase constant.
First, we need to find the values of A, ω, and φ.
Given:
- Amplitude (initial displacement): x0 = 1.7 m
- Angular frequency: ω = √(k/m)
(where k is the spring constant and m is the mass)
- Phase constant: φ = 0 (since the block starts from its maximum displacement and is released with an initial speed)
Plugging in the values, we have:
x(t) = 1.7 * cos(√(k/m) * t)
2. How long will it take for the mass to first return to the equilibrium position?
The mass will first return to the equilibrium position when x(t) = 0.
So, we can set up the equation:
0 = 1.7 * cos(√(k/m) * t)
To solve for t, we rearrange the equation:
cos(√(k/m) * t) = 0
Using the property of cosine function, we know that cos(0) = 1, and cos(π) = -1 (where π is pi).
Therefore, we can set √(k/m) * t = π (or any odd multiple of π) to get the first return to the equilibrium position.
Solving for t, we have:
√(k/m) * t = π
t = π * √(m/k)
3. How long will it take for the spring to first become completely extended?
The spring will first become completely extended when |x(t)| = |A|.
Using the equation x(t) = 1.7 * cos(√(k/m) * t), we can set |x(t)| = |A| to find the time at which the spring becomes completely extended.
|1.7 * cos(√(k/m) * t)| = |1.7|
Simplifying:
cos(√(k/m) * t) = 1
Using the property of the cosine function, we know that cos(0) = 1.
Therefore, we can set √(k/m) * t = 0 to get the first complete extension of the spring.
Solving for t, we have:
√(k/m) * t = 0
t = 0
This means that the spring is already completely extended at the initial time of release.
To summarize:
(1) The position of the block as a function of time is x(t) = 1.7 * cos(√(k/m) * t).
(2) The time it will take for the mass to first return to the equilibrium position is t = π * √(m/k).
(3) The time it will take for the spring to first become completely extended is t = 0 (already extended at the initial release).