Explanation:
Given that,
Mass of the block, m = 300 g = 0.3 kg
Linear spring constant, k = 6.5 N/m
(a) Let T is the period of the block's motion. It is given by :
[tex]T=\dfrac{2\pi}{\omega}[/tex]
where
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]= angular frequency
Also, [tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}[/tex]
[tex]T=\dfrac{2\pi}{\sqrt{\dfrac{6.5}{0.3}}}[/tex]
T = 1.34 seconds
(b) The maximum acceleration of the block is, [tex]a_{max}=2\ m/s^2[/tex]
The maximum acceleration is given by :
[tex]a_{max}=\omega^2A[/tex]
A is the amplitude of the motion,
[tex]A=\dfrac{a_{max}}{\omega^2}[/tex]
[tex]A=\dfrac{a_{max}}{(\sqrt{\dfrac{k}{m}})^2}[/tex]
[tex]A=\dfrac{ma_{max}}{k}[/tex]
[tex]A=\dfrac{0.3\times 2}{6.5}[/tex]
A = 0.09 meters
Hence, this is the required solution.
