Answer:
Explanation:
Given
Mechanical Energy of Spring-Block system is 1 J
Maximum Amplitude is [tex]A=10 cm[/tex]
maximum speed [tex]v_{max}=1.2 m/s[/tex]
Suppose [tex]x=A\sin \omega t [/tex]be general equation of motion of spring-mass system
where A=max amplitude
[tex]\omega [/tex]=Natural frequency of oscillation
t=time
[tex]v_{max}=A\omega =1.2[/tex]
[tex]0.1\cdot \omega =1.2[/tex]
[tex]\omega =12 rad/s[/tex]
maximum kinetic Energy must be equal to total Mechanical Energy when spring is un deformed i.e. at starting Position
[tex]\frac{1}{2}mv_{max}^2=1[/tex]
[tex]m=\frac{2}{1.2^2}=\frac{2}{1.44}=1.38 kg[/tex]
Also [tex]\omega [/tex]is also given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]k=\omega ^2\cdot m[/tex]
where k= spring constant
[tex]k=12^2\cdot 1.38 [/tex]
[tex]k=200 N/m[/tex]
