Answer:
a)TE=0.0245 J
b)A = 0.0128 m
c)V=0.57 m/s
Explanation:
Given that
m = 0.150 kg
K= 300 N/m
x= 0.012 ,v= 0.2 m/s
The velocity of the toy at any point given as
[tex]v=\omega\sqrt{A^2-x^2}[/tex]
[tex]\omega=\sqrt{\dfrac{K}{m}}[/tex]
[tex]v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}[/tex]
[tex]0.2=\sqrt{\dfrac{300}{0.15}}\times \sqrt{A^2-0.012^2}[/tex]
2 x 10⁻⁵ = A² - 0.000144
A=0.0128 m
Amplitude ,A = 0.0128 m
The total energy TE
[tex]TE=\dfrac{1}{2}KA^2[/tex]
[tex]TE=\dfrac{1}{2}300\times 0.0128^2[/tex]
TE=0.0245 J
The maximum speed
[tex]V=\omega A[/tex]
[tex]V=\sqrt{\dfrac{K}{m}}\times A[/tex]
[tex]V=\sqrt{\dfrac{300}{0.15}}\times 0.0128[/tex]
V=0.57 m/s
