Question

A mass of 0.2kg is suspended from a fixed
point by a light spring the mass is
extended from 0.02m from the equilibrium
position. The mass is then pulled down
by 9.03m and released find the time
period and maximum kinetic energy of the mass.​

Answer 1

Answer:the time when the particle is at the mid period of it its total distance

Explanation:

Answer 2

Time period of mass and maximum kinetic energy is 0.28 s and 4077.045 J respectively.

Explanation:

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Given that

mass , m = 0.2 kg

initial extension , x= 0.02 m

Final ,A= 9.03 m

Lets take spring constant = k

m g = k x

$k=\dfrac{0.2\times 10}{0.02}\ N/m$

k=100 N/m

Time period

$T=2\pi \sqrt{\dfrac{m}{k} }$

$T=2\pi \times \sqrt{\dfrac{0.2}{100} }$

T=0.28 s

The maximum kinetic energy

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2$

v=velocity

$KE=\dfrac{1}{2}kA^2$

$KE=\dfrac{1}{2}100\times 9.03^2\ J$

KE=4077.045 J

Therefore the time period and kinetic energy will be 0.28 s and 4077.045 J respectively.