A 1.65kg mass stretches a vertical spring 0.215m. If the spring is stretched an additional 0.130m released, how long does it take to reach the new equillibrium position again?
To answer this question, we need to use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The equation for Hooke's law is:
F = -kx
Where:
F is the force exerted by the spring (in Newtons),
k is the spring constant (in N/m),
x is the displacement of the spring from its equilibrium position (in meters).
We can rearrange this equation to solve for the spring constant:
k = -F/x
First, we need to calculate the spring constant using the given information. In this case, the mass of 1.65kg stretches the spring by 0.215m. We can calculate the force exerted by the spring using the equation:
F = mg
Where:
m is the mass of the object (in kg),
g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values:
F = (1.65 kg) * (9.8 m/s²) ≈ 16.17 N
Now we can calculate the spring constant:
k = -(16.17 N) / (0.215 m) ≈ -75.27 N/m)
Now that we know the spring constant, we can calculate the time it takes for the spring-mass system to reach the new equilibrium position after releasing the additional 0.130m of stretch.
The formula to calculate the period (T) of motion for a mass-spring system is:
T = 2π * sqrt(m/k)
Where:
T is the time period (in seconds),
m is the mass of the object (in kg),
k is the spring constant (in N/m).
Plugging in the values:
T = 2π * sqrt(1.65 kg / 75.27 N/m) ≈ 0.955 seconds
Therefore, it would take approximately 0.955 seconds for the spring-mass system to reach the new equilibrium position again after being released from the additional 0.130m of stretch.