A 2 kg block of cheddar cheese hangs from an ideal spring of negligible mass. When displaced from equilibrium and released, the cheese oscillates with a period of 0.4 secs. How much the spring stretched when the cheese hangs in equilibrium?
period=2PIsqrt(weight/springconstant)
solve for spring constant.
then weight= kx solve for x.
To determine how much the spring stretched when the cheese hangs in equilibrium, we need to find the restoring force exerted by the spring when the cheese is displaced from equilibrium.
The period of oscillation of an object attached to a spring can be calculated using the formula:
T = 2π√(m/k)
Where T is the period, m is the mass of the object, and k is the spring constant.
Rearranging the formula to solve for the spring constant:
k = (4π²m) / T²
Substituting the given values:
m = 2 kg
T = 0.4 secs
k = (4π² * 2 kg) / (0.4 secs)²
k = (4π² * 2 kg) / (0.16 secs²)
k = 157.93 N/m (rounded to two decimal places)
The spring constant is 157.93 N/m.
The equilibrium position occurs when the restoring force exerted by the spring is equal in magnitude and opposite in direction to the weight of the object hanging from it. Therefore, we can use Hooke's Law to find the spring displacement at equilibrium:
F_spring = k * x
Where F_spring is the force exerted by the spring, k is the spring constant, and x is the spring displacement from equilibrium.
Setting the force exerted by the spring equal to the weight of the cheese (mg):
mg = k * x
Rearranging the formula to solve for x:
x = (mg) / k
Substituting the given values:
m = 2 kg
g = 9.8 m/s²
k = 157.93 N/m
x = (2 kg * 9.8 m/s²) / 157.93 N/m
x = 0.126 m (rounded to three decimal places)
Therefore, the spring stretched approximately 0.126 meters when the cheese hangs in equilibrium.
To find the amount the spring stretched when the cheese hangs in equilibrium, we need to use the formula for the period of oscillation of a mass-spring system.
The formula for the period is:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
We are given the mass of the cheese, which is 2 kg, and the period of oscillation, which is 0.4 seconds. We need to solve for the spring constant (k).
Rearranging the formula, we get:
T^2 = (4π^2)(m/k)
Plugging in the known values, we have:
(0.4^2) = (4π^2)(2/k)
Simplifying further, we get:
0.16 = (8π^2)/k
To isolate k, we can rearrange the equation:
k = (8π^2) / 0.16
k ≈ 157.92 N/m
Now that we have the spring constant, we can find the amount the spring stretched at equilibrium.
According to Hooke's Law, the force exerted by the spring is proportional to the displacement from equilibrium. The equation is:
F = -kx
where F is the force, k is the spring constant, and x is the displacement.
Since the cheese hangs in equilibrium, the net force acting on it is zero. Therefore, the force exerted by the spring is equal to the weight of the cheese.
F = mg
where m is the mass and g is the acceleration due to gravity.
Plugging in the values, we have:
F = (2 kg)(9.8 m/s^2)
F = 19.6 N
Setting this equal to the force from Hooke's Law:
19.6 N = -kx
Solving for x, the displacement, we get:
x = -19.6 N / (-157.92 N/m)
x ≈ 0.124 meters
Therefore, the spring stretches approximately 0.124 meters when the cheese hangs in equilibrium.