Answer:
The amplitude of the spring is 32.6 cm.
Explanation:
It is given that,
Mass of the block, m = 2 kg
Force constant of the spring, k = 300 N/m
At t = 0, the velocity of the block, v = -4 m/s
Displacement of the block, x = 0.2 mm = 0.0002 m
We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :
[tex]v=\omega\sqrt{A^2-x^2}[/tex]
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{300}{2}}[/tex]
[tex]\omega=12.24\ rad/s[/tex]
So, [tex]\dfrac{v^2}{\omega^2}+x^2=A^2[/tex]
[tex]\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2[/tex]
A = 0.326 m
or
A = 32.6 cm
So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.
