A 200 gram mass is attached to a horizontal spring whose other end is connected

Question

A 200 gram mass is attached to a horizontal spring whose other end is connected to the wall. The mass is pulled to right and released a distance of 25 cm from equilibrium. At time t=0 s it passes through the equilibrium position, headed to the left. It takes the mass 0.55 seconds (after t=0 s) to return, for the first time, to the equilibrium position. i) What is the value of the spring constant? ii) What is the total mechanical energy of the system? iii) What is the magnitude of the maximum acceleration of the mass? At what position(s) does this occur?

Explanation / Answer

Here ,

m = 0.20 Kg

d = 0.25 m

t = 0.55 s

i) for the value of the spring constant is k

4 * 0.55 = 2pi * sqrt(0.20/k)

solving for k

k = 1.63 N/m

the spring constant is 1.63 N/m

ii)

total mechanical energy of system = 0.50 * 1.63 * 0.25^2

total mechanical energy of system = 0.0509 J

total mechanical energy of system is 0.0509 J

iii)

magnitude of acceleration = 1.63 * 0.25/.20

magnitude of acceleration = 2.04 m/s^2

it occurs at the maximum positions

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