Why does the equilibrium position of the spring change when a mass is added to the spring? will the mass oscillate around the new equilibrium position of the spring or the previous position without a mass attached to the spring? if the equilibrium position of the spring changes by 20 cm (assuming no initial mass) when a mass is added to the spring with constant 4.9 kg/s^2, what is the mass of the object attached to the spring?

whenever a new mass is added to spring the equilibrium state of the spring will change and then the mas will start oscillating about the new equilibrium position.

So here we can say that mass will always oscillate about its new equilibrium position if it is changed

Now in next situation it is given that

spring constant(k) = 4.9 kg/s^2

displacement of mass after suspending through spring = 20 cm

now we have the formula of spring force as

here applied force is due to gravity so we have

so mass of object will be 100 g or 0.10 kg

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.