A 100-coil spring has a spring constant of 760 N/m. It is cut into four shorter springs, each of which has 25 coils. One end of a 25-coil spring is attached to a wall. An object of mass 45 kg is attached to the other end of the spring, and the system is set into horizontal oscillation. What is the angular frequency of the motion?

Answer 1

The formula for the angular frequency of an oscillating spring is given by:

ω = sqrt(k/m)

where:
ω = angular frequency
k = spring constant
m = mass attached to the spring

First, let's find the spring constant of the 25-coil spring. Since the original spring has 100 coils, and the spring constant is directly proportional to the number of coils, the spring constant of the 25-coil spring can be calculated as follows:

k1 = k * (n1/n2)^2

where:
k1 = spring constant of the 25-coil spring
k = spring constant of the original 100-coil spring
n1 = number of coils in the 25-coil spring
n2 = number of coils in the original 100-coil spring

Using the formula, we can substitute the given values:

k1 = 760 N/m * (25/100)^2
k1 = 760 N/m * (1/16)
k1 = 47.5 N/m

Next, we substitute the values into the formula for angular frequency:

ω = sqrt(k1/m)
ω = sqrt(47.5 N/m / 45 kg)
ω = sqrt(1.0556 N/kg)
ω ≈ 1.03 rad/s

Therefore, the angular frequency of the motion is approximately 1.03 rad/s.