Answer:
Explanation:
Given
mass of block [tex]m=2.12 kg[/tex]
spring constant [tex]k=250 n/m[/tex]
speed of block at t=0 is [tex]v=13 m/s[/tex]
natural Frequency of oscillation [tex]\omega _n=\sqrt{\frac{k}{m}}[/tex]
[tex]\omega _n=\sqrt{\frac{k}{m}}[/tex]
[tex]\omega _n=\sqrt{\frac{250}{2.12}}[/tex]
[tex]\omega _n=\sqrt{117.92}[/tex]
[tex]\omega _n=10.85 rad/s[/tex]
Suppose [tex]x=A\cos (\omega _nt+\phi )[/tex]
where [tex]\phi =Phase\ difference[/tex]
at t=0 spring is neither compressed
Amplitude is given by
[tex]A=\sqrt{x^2+(\frac{v}{\omega })^2}[/tex]
at t=0 x=0 i.e. at mean Position
[tex]A=\sqrt{0+(1.198)^2}[/tex]
[tex]A=1.198 m[/tex]
at [tex]t=0[/tex]
[tex]x=A\cos (\omega _nt+\phi )[/tex]
[tex]x=1.198\cos (0+\phi )[/tex]
[tex]0=\cos (\phi )[/tex]
therefore [tex]\phi =\frac{\pi }{2}[/tex]
