Answer:
A)A=0.075 m
B)v= 0.21 m/s
Explanation:
Given that
m = 0.75 kg
K= 13.5 N
The natural frequency of the block given as
[tex]\omega =\sqrt{\dfrac{K}{m}}[/tex]
The maximum speed v given as
[tex]v=\omega A[/tex]
A=Amplitude
[tex]v=\sqrt{\dfrac{K}{m}}\times A[/tex]
[tex]0.32=\sqrt{\dfrac{13.5}{0.75}}\times A[/tex]
A=0.075 m
A= 0.75 cm
The speed at distance x
[tex]v=\omega \sqrt{A^2-x^2}[/tex]
[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]
[tex]v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}[/tex]
v= 0.21 m/s
