Question:

A mass-spring system undergoes periodic motion. The spring constant is 40 N/m and the mass attached to the spring is 0.5 kg. The initial displacement of the mass from the equilibrium position is 0.1 m. The amplitude is defined as the maximum displacement from equilibrium.

(a) Calculate the angular frequency of the mass-spring system.

(b) Determine the equation of motion for the mass-spring system.

(c) Calculate the period and frequency of the periodic motion.

(d) Calculate the maximum velocity and acceleration of the mass during its motion.

(e) Sketch a displacement-time graph for one complete cycle of the mass-spring system.

Answer:

(a) The angular frequency of a mass-spring system can be calculated using the formula:

Angular frequency (ω) = √(k/m)

Where: k = spring constant = 40 N/m m = mass = 0.5 kg

Plugging in the given values:

ω = √(40/0.5) = √(80) = 8.94 rad/s

Therefore, the angular frequency of the mass-spring system is 8.94 rad/s.

(b) The equation of motion for a mass-spring system can be expressed as:

x(t) = A * cos(ω * t + φ)

Where: x(t) = displacement from equilibrium position at time t A = amplitude of the motion ω = angular frequency t = time φ = phase angle

Given: Amplitude (A) = 0.1 m Angular frequency (ω) = 8.94 rad/s

The equation of motion becomes:

x(t) = 0.1 * cos(8.94 * t + φ)

(c) The period (T) of the motion is the time taken for one complete cycle. It can be calculated using the formula:

Period (T) = 2π / ω

Plugging in the given value of ω:

T = 2π / 8.94 = 0.706 s

The frequency(f) of the motion is the number of cycles per unit time and is given by:

Frequency (f) = 1 / T

Plugging in the value of T:

f = 1 / 0.706 = 1.419 Hz

Therefore, the period of the periodic motion is 0.706 s and the frequency is 1.419 Hz.

(d) The maximum velocity (v_max) and acceleration (a_max) of the mass during its motion can be determined using the formulas:

Maximum velocity (v_max) = A * ω Maximum acceleration (a_max) = A * ω^2

Given: Amplitude (A) = 0.1 m Angular frequency (ω) = 8.94 rad/s

Plugging in the values:

v_max = 0.1 * 8.94 = 0.894 m/s

a_max = 0.1 * (8.94)^2 = 7.961 m/s^2

Therefore, the maximum velocity of the mass during its motion is 0.894 m/s and the maximum acceleration is 7.961 m/s^2.

(e)

The above graph represents a displacement-time graph for one complete cycle of the mass-spring system. The oscillation starts at the maximum positive displacement (0.1 m), then decreases to zero at the equilibrium position, reaches the maximum negative displacement (-0.1 m), returns to zero at the equilibrium position, and completes one full cycle.