Answer:11.55 N
Explanation:
Given
mass [tex]m=0.38 kg[/tex]
Spring constant [tex]k=106 N/m[/tex]
initial compression [tex]x=0.109 m[/tex]
mass will perform simple harmonic motion with general equation of motion given by
[tex]x=A\sin \omega t[/tex]
where [tex]x=Position\ of\ Mass [/tex]
[tex]A=maximum\ Amplitude[/tex]
[tex]\omega =Natural\ frequency\ of\ oscillation=\sqrt{\frac{k}{m}}[/tex]
[tex]\omega =16.70 rad/s[/tex]
[tex]t=time [/tex]
therefore acceleration is given by
[tex]a=-A\omega ^2\sin \omega t[/tex]
[tex]a_{max}=A\omega ^2[/tex]
initial compression is equivalent to max amplitude
thus [tex]A=0.109 m[/tex]
[tex]a_{max}=A\omega ^2=0.109\times (16.70)^2=30.405 m/s^2[/tex]
therefore Maximum Force [tex]F=ma_{max}[/tex]
[tex]F=0.38\times 30.405=11.55 N[/tex]
