Respuesta :
Explanation:
Given that,
Mass of the block, m = 5 kg
Spring constant, k = 2000 N/m
The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium.
(a) The angular frequency is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Since, [tex]\omega=2\pi f[/tex], f is frequency
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{2000}{5}} \\\\f=3.18\ Hz[/tex]
(b) The velocity of particle executing SHM is given by :
[tex]v=\omega\sqrt{A^2-x^2}[/tex]
x is displacement from equilibrium position
[tex]\dfrac{v}{\omega}=\sqrt{A^2-x^2} \\\\(\dfrac{v}{\omega})^2=A^2-x^2\\\\A^2=(\dfrac{v}{\omega})^2+x^2\\\\A=\sqrt{(\dfrac{v}{\omega})^2+x^2} \\\\A=\sqrt{(\dfrac{1}{2\pi \times 3.18})^2+(0.05)^2}\\A=0.070\ m\\\\A=7\ cm[/tex]
(c) The total mechanical energy of the motion in SHM is given by :
[tex]E=\dfrac{1}{2}kA^2\\\\E=\dfrac{1}{2}\times 2000\times (0.070)^2\\\\E=4.9\ J[/tex]
Hence, this is the required solution.
