You hang mass m=0.22 kg on a spring with constant 12 N/m. You pull the mass down a distance of 0.04 m from its equilibrium position and release it. You measure the period to be 0.85 s.
If you instead pull the mass down 0.12 m from its equilibrium position and release it, what would the new period be?
I believe the period time will be tripled from the original time at 2.55s because the distance that the spring was puled down was 3 times the original length.
Answer= 2.55 seconds
The period is
T= 2π•sqrt(m/k) =
=2•3.14•sqrt(0.22/12) =0.85 s.
It doesn’t depend on „x“
To find the new period, we need to use Hooke's law and the equation for the period of a mass-spring system.
Hooke's law states that the force exerted by a spring is directly proportional to the displacement from the equilibrium position. Mathematically, it can be written as:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the spring constant is given as k = 12 N/m, and the displacement is x = 0.04 m.
So, the force exerted by the spring is:
F = -(12 N/m) * (0.04 m)
F = -0.48 N
Now, let's calculate the angular frequency (ω) using the equation:
ω = √(k/m)
where k is the spring constant and m is the mass.
ω = √(12 N/m / 0.22 kg)
ω = √(54.55 rad/s^2)
ω ≈ 7.39 rad/s
Next, let's calculate the new displacement (x) for the second scenario, where x = 0.12 m.
Now, let's calculate the new period (T) using the formula:
T = 2π / ω
T = 2π / 7.39 rad/s
T ≈ 0.85 s
Therefore, the new period when you pull the mass down 0.12 m from its equilibrium position and release it would also be approximately 0.85 seconds.
To find the new period, we need to use the equation for the period of a mass-spring system, which is given by:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the spring constant.
Given that the mass is m=0.22 kg and the spring constant is k=12 N/m, we can substitute these values into the equation:
T = 2π√(0.22/12)
Now let's calculate the original period:
T = 2π√(0.22/12)
T ≈ 2π√0.0183
T ≈ 2π√0.0183
T ≈ 2π(0.135)
The period T is approximately 0.85 seconds, as stated in the problem.
Now let's consider the situation when we pull the mass down 0.12 m from its equilibrium position. To find the new period, we need to calculate the new value of the period using the new displacement.
T' = 2π√(m/k')
Where T' is the new period, m is the mass, and k' is the new spring constant when the displacement is changed to 0.12 m.
We know that the spring constant k is given by k = mω^2, where ω is the angular frequency. The angular frequency ω is related to the period T by the equation ω = 2π/T.
Plugging this equation into the expression for k gives us k = m(2π/T)^2.
Now we can calculate the new spring constant:
k' = m(2π/T)^2
k' = 0.22(2π/0.85)^2
k' ≈ 0.22(25.28)^2
Now, we can substitute this new spring constant into the formula for the period:
T' = 2π√(0.22/k')
T' = 2π√(0.22/(0.22(25.28)^2))
T' = 2π√(1/(25.28)^2)
T' = 2π/(25.28)
Calculating the value:
T' ≈ 0.785 s
So, the new period when the mass is pulled down 0.12 m from its equilibrium position is approximately 0.785 seconds.