Explanation:
Given that,
Mass, m = 343 g = 0.343 kg
Force constant, k = 3 N/m
Displacement in the mass from equilibrium position is 4 cm
(a) The time period of the mass that oscillates is given by :
[tex]T=2\pi \sqrt{\dfrac{m}{k}} \\\\T=2\pi \sqrt{\dfrac{0.343}{3}} \\\\T=2.12\ s[/tex]
(b) Firstly lets find the angular frequency. It is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{3}{0.343}} \\\\\omega=2.95\ rad/s[/tex]
Now the maximum speed of the mass in SHM is given by :
[tex]v=A\omega\\\\v=0.04\times 2.95\\\\v=0.118\ m/s[/tex]
(c) The maximum acceleration of the mass is given by :
[tex]a=A\omega^2\\\\a=0.04\times (2.95)^2\\\\a=0.3481\ m/s^2[/tex]
Hence, this is the required solution.
