Explanation:
Given that,
Length of the spring, l = 50 cm = 0.5 m
Mass connected to the end, m = 410 g = 0.41 kg
The mass is released and falls, stretching the spring by 12 cm before coming to rest at its lowest point.
(a) At equilibrium, the weight of the object attached is balanced by 10 cm below the release point such that,
kx = mg
k is spring constant of the spring
[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{0.41\times 9.8}{0.12}\\\\k=33.48\ N/m[/tex]
(b) Amplitude of the oscillation is half of the total distance traveled. So, it is 6 cm.
(c) The frequency of the oscillation is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
Putting all the values
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{33.48}{0.41}} \\\\f=1.43\ Hz[/tex]
Hence, the spring constant of the spring is 33.48 N/m, the amplitude of the oscillation is 6 cm and the frequency of the oscillation is 1.43 Hz.
