A 55.0 kg circus performer oscillates up and down at the end of a long elastic rope at a rate of once every 7.20 s. The elastic rope obeys Hooke's Law. By how much is the rope extended beyond its unloaded length when the performer hangs at rest?
The period of oscillations is
T = 2•π•sqr(m/k)
k= 4• π^2•m/T^2 =41.84 N/m
Hook’s law: F = - kx, F=-mg
mg = kx,
x=mg/k=55•9.8/41.84 =12.88 m
Thank you , Elena.
Thank you!
To find the amount by which the rope is extended beyond its unloaded length when the performer hangs at rest, we need to use Hooke's Law.
Hooke's Law states that the force required to extend or compress a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be expressed as:
F = -kx
where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the performer is hanging at rest, so the net force on the rope is zero. This means that the force pulling the performer up is balanced by the force pulling the performer down.
Since the force upward is equal to the force downward, we can set up the equation:
kx = mg
where k is the spring constant, x is the displacement of the rope, m is the mass of the performer, and g is the acceleration due to gravity.
We can rearrange this equation to solve for x:
x = mg / k
To find the spring constant, we need to use the formula for the period of an oscillating spring:
T = 2π √(m / k)
where T is the period of oscillation. Rearranging this equation, we get:
k = (4π²m) / T²
Now we can substitute the values given in the problem:
m = 55.0 kg
T = 7.20 s
First, let's find the spring constant, k:
k = (4π² * 55.0) / (7.20)² ≈ 209.1 N/m
Now, let's find the displacement, x:
x = (55.0 kg * 9.8 m/s²) / 209.1 N/m ≈ 2.61 m
Therefore, the rope is extended beyond its unloaded length by approximately 2.61 meters when the performer hangs at rest.