Explanation:
Given that,
Mass of an object, m = 250 g = 0.25 kg
Spring constant, k = 48 N/m
The amplitude of the oscillation, A = 5.42 cm = 0.0542 m
1. At equilibrium,
ma = kx
Where
a is the acceleration of the object
So,
[tex]a=\dfrac{kx}{m}\\\\a=\dfrac{48\times 0.0542}{0.25}\\\\a=10.4\ m/s^2[/tex]
2. The maximum speed of the object is :
[tex]v=A\omega\\\\v=A\sqrt{\dfrac{k}{m}}\\\\v=0.0542\times \sqrt{\dfrac{48}{0.25}}\\\\v=0.75\ m/s[/tex]
Hence, this is the required solution.
