1) A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. What is the speed of the 1.0 kg mass at the equilibrium position?
At the equilibrium position, all of the initial stored potential energy changes to KE.
1/2 m v^2=1/2 k (.1)^2
solve for v
Nil
To find the speed of the 1.0 kg mass at the equilibrium position, we need to understand the properties and behavior of ideal springs in simple harmonic motion.
Simple harmonic motion is a periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction. In the case of an ideal spring, Hooke's Law describes this relationship as F = -kx, where F is the restoring force, k is the force constant (also known as the spring constant), and x is the displacement from the equilibrium position.
Given:
Mass (m) = 1.0 kg
Force constant (k) = 400 N/m
Amplitude (A) = 10 cm = 0.1 m
First, let's determine the angular frequency (ω) of the simple harmonic motion. The angular frequency is defined as ω = √(k/m), where √ stands for the square root.
ω = √(k/m) = √(400 N/m / 1.0 kg) = √400 rad/s = 20 rad/s
The velocity (v) of an object undergoing simple harmonic motion can be expressed as v = ωA, where A is the amplitude.
v = ωA = (20 rad/s) * (0.1 m) = 2 m/s
Thus, the speed of the 1.0 kg mass at the equilibrium position is 2 m/s.