Anvular frequency of syatem
w^2 = k/m
w^2 = 250 / 0.1
w = 50 rad/s
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a)
w = 2"pi / T
50 = 2 pi / T
T = 0.1256 s
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b)
Instantaneous position
x = A cos ( wt)
x = 0.4 cos ( 50 t)
At t = 0.045 s
x = - 0.251 m
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c)
Mechanical energy
E= 0. 5 k x^2 = 20 J
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d)
Using energy conservation
0.5 k x^2 = 0.5 k a^2 - 0.5 m v^2
250 x^2 = 250* 0.4^2 - 0.1* 5^2
x = ± 0.387 m
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Do comment in case any doubt will reply for sure, goodluck
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