Question

2. A spring with constant 1.46 N/m has an unknown mass attached to it. It is pulled a set distance and released from rest. The resulting graph for position of the unknown mass as a function of time is shown below. Oscillating Mass-Spring System 1 position (m) 0.8 0.6 04 02 0 -0.2 5 -0.4 -0.8 times) a) What is the frequency? (1 point) b) What is the amplitude? (1 point) c) What is the angular frequency? (1 points) d) What is the mass being oscillated? (1 point) e) What is the maximum velocity? (2 points) f) What is the maximum acceleration? (2 points) g) Calculate the total mechanical energy in the spring-mass system. (2 points) h) Write equation that describes the acceleration of the system. (2 points)

Answer 1

Answer:

Given, spring constant k = 1.46 N/m

From the given x-t graph,

Time period is T = 4.5 s

(a) Frequency f = 1/T = 1/(4.5 s) = 0.222 Hz

(b) Amplitude is nothing but the maximum displacement of the oscillating object. From the graph, the amplitude is

A = 0.9 m

(c) Angular frequency $\omega$ = 2$\pi$f = 2$\pi$(0.222 Hz) = 1.394 rad/s

(d) The relation between spring constant, mass and angular frequency is k = m$\omega$²

Thus, mass m = k/$\omega$² = (1.46 N/m) / (1.394 rad/s)² = 0.751 kg

(e) The expression for the maximum velocity is v_max = $\omega$ A

Thus, v_max = (1.394 rad/s)(0.9 m) = 1.254 m/s

(f) Maximum acceleration is a_max = $\omega$ ²A = (1.394 rad/s)² (0.9 m) = 1.748 m/s²

(g) The total mechanical energy of the system is E = 1/2 kA² = 1/2 (1.46 N/m) (0.9 m)² = 0.591 J

(h) Using the magnitude of the elastic force F = kx and

the Newton's second law, F = ma

Equate these two equations, the we obtain,

    kx = ma

Therefore, acceleration a = kx/m