Answer:
a)V= 2.09 m/s
b)TE= 0.56 J
c)K=50.47 N/m
d)a=29.23 m/s²
Explanation:
Given that
m = 0.26 kg , T= 0.45 s ,A= 0.15 m
We know that time period given as
[tex]T=\dfrac{2\pi}{\omega}[/tex]
ω =Angular frequency
[tex]{\omega}=\dfrac{2\pi}{T}[/tex]
[tex]{\omega}=\dfrac{2\pi}{0.45}[/tex]
ω = 13.96 rad/s
The velocity at equilibrium
V= ω A
V= 13.96 x 0.15
V= 2.09 m/s
The total energy TE
[tex]TE=\dfrac{1}{2}mV^2[/tex]
[tex]TE=\dfrac{1}{2}\times 0.26\times 2.09^2[/tex]
TE= 0.56 J
The spring constant K
Maximum stored energy in the spring
[tex]U=\dfrac{1}{2}KA^2[/tex]
From energy balance
U= TE
[tex]\dfrac{1}{2}KA^2=\dfrac{1}{2}mV^2[/tex]
K A² = m V²
[tex]=\dfrac{mV^2}{A^2}[/tex]
[tex]K=\dfrac{0.26\times 2.09^2}{0.15^2}[/tex]
K=50.47 N/m
The maximum acceleration a
a= ω² A
a = 13.96² x 0.15 m/s²
a=29.23 m/s²
