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 apply the principles of simple harmonic motion. Here's how we can solve them 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 obtained using the equation 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
- φ is the phase constant
To find the values of A, ω, and φ, we need to consider the given information about the system.
The amplitude of the motion can be determined by the initial compression of the spring:
A = x₀ = 1.7 m
The angular frequency can be calculated using the formula:
ω = sqrt(k/m)
Where:
- k is the spring constant (20 N/m)
- m is the mass of the block (6 kg)
Substituting the values, we get:
ω = sqrt(20/6) ≈ 1.825 rad/s
The phase constant φ can be found by considering the initial conditions. At t = 0, the block is released from the compressed position with an initial speed v₀ = 5 m/s towards the equilibrium. We can determine φ based on the initial velocity:
v₀ = -A * ω * sin(φ)
Solving for sin(φ):
sin(φ) = -v₀ / (A * ω)
Substituting the known values:
sin(φ) = -5 / (1.7 * 1.825)
Now, we have all the necessary values to find the position of the block as a function of time.
(2) How long will it take for the mass to first return to the equilibrium position?
To determine the time it takes for the mass to return to the equilibrium position, we need to find the period of the motion.
The period of simple harmonic motion is given by the formula:
T = 2π / ω
Where:
- T is the period
- ω is the angular frequency
Substituting the value of ω, we get:
T = 2π / 1.825
The time it takes for the mass to return to the equilibrium position will be equal to half the period (T/2).
(3) How long will it take for the spring to first become completely extended?
To find the time it takes for the spring to become completely extended, we need to consider the amplitude of the motion.
The maximum displacement of the spring from its equilibrium position is the amplitude (A). The spring will become completely extended when its displacement is equal to the original length of the spring (l₀).
So, the time it takes for the spring to become completely extended is equal to the time it takes for the block to travel a distance equal to l₀.
Using the equation of motion for simple harmonic motion, we can find the time it takes for the block to travel a distance equal to l₀.
Hope this helps! Let me know if you have any more questions.