A spring stretches 5.00 cm when a 10.00 g object is hung from it. The object is replaced with another object of mass 20.00 g that oscillates in simple harmonic motion.
a) Calculate the spring constant in N/m. Use 9.8 m/s2 for g.
b) Calculate the period of motion in seconds.
weight = .010*9.8 = .098 Newtons
x = .05 m
k = 1.96 N/m
omega = sqrt (k/m) = 2 pi f
so
f = 2.23
T = 1/f = .45 seconds
a) To calculate the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is: F = k * x
Where:
F is the force applied to the spring (in Newtons),
k is the spring constant (in N/m),
and x is the displacement of the spring (in meters).
In this case, we are given:
x = 0.0500 m (5.00 cm) (displacement)
m = 0.0100 kg (mass)
g = 9.8 m/s² (acceleration due to gravity)
Since the object is hanging vertically, the force applied to the spring is the weight of the object, given by:
F = m * g
Substituting these values into the equation, we have:
m * g = k * x
Solving for k:
k = (m * g) / x
Substituting the given values, we can calculate:
k = (0.0100 kg * 9.8 m/s²) / 0.0500 m
Calculating this expression:
k = 1.96 N/m
Therefore, the spring constant is 1.96 N/m.
b) The period of motion for an object undergoing simple harmonic motion with a spring can be calculated using the formula:
T = 2π * √(m / k)
Where:
T is the period of motion (in seconds),
π is a mathematical constant approximately equal to 3.14,
m is the mass of the object (in kg),
and k is the spring constant (in N/m).
Substituting the given values, we can calculate:
T = 2π * √(0.0200 kg / 1.96 N/m)
Calculating this expression:
T = 0.898 seconds
Therefore, the period of motion is approximately 0.898 seconds.