Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.
[tex]w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s[/tex]
The maximum velocity is
[tex]v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s[/tex]
