The 4 kg mass attached to a 10 N/m spring oscillates with a natural frequency of 1.58 rad/s and a period of 4 seconds. Displacing it 0.4 meters and releasing it from rest sets the amplitude at 0.4 meters. Releasing it from equilibrium with an initial 0.7 m/s velocity increases the amplitude to 0.54 meters and its maximum velocity to 0.94 m/s.
System parameters:
Mass (m) = 4 kg
Spring constant (k) = 10 N/m
Natural frequency (ω):
ω = √(k / m) = √(10 N/m / 4 kg) = 1.581 rad/s
Period (T):
T = 2π / ω = 2π / 1.581 rad/s ≈ 3.974 s
Case 1: Displaced 0.4 m, released from rest
The initial displacement acts as the amplitude (A) of the motion:
A₁ = 0.4 m
Case 2: Released from equilibrium with initial velocity (v₀) = 0.7 m/s
Total mechanical energy is conserved, so the initial kinetic energy (KE₁ = 1/2 mv₀²) is converted to potential energy (PE = 1/2 kA²).
A₂ = √(A₁² + (v₀² / ω²)²) = √(0.4² + (0.7² / 1.581²)²) ≈ 0.539 m
Case 3: Displaced 0.4 m, released with v₀ = 0.7 m/s
Similar to Case 2, but both initial displacement and velocity contribute to the amplitude:
A₃ = √(A₁² + (v₀² / ω²)²) ≈ 0.597 m
Maximum velocity (vmax) in Case 3:
vmax = ω * A₃ ≈ 0.943 m/s
Question:-
4 kg mass is attached to spring with spring constant Nt/m:
What is the frequency of the simple harmonic motion? radians/second What is the period? seconds Suppose the mass is displaced 0.4 meters from its equilibrium position and released from rest: What is the amplitude of the motion?
meters
Suppose the mass is released from the equilibrium position with an initial velocity of 0.7 meters/sec: What is the amplitude of the motion?
meters
Suppose the mass is is displaced 0.4 meters from the equilibrium position and released with an initial velocity of 0.7 meters/sec: What is the amplitude of the motion? meters What is the maximum velocity? m/s