Question

A vertically suspend spring attached to mass M is held at natural length by a support beneath the mass.
How would you (not experimentally) measure the amplitude of the oscillation? assuming no damping, and ideal spring

If I know at the new equilibrium the spring is extended by a metres from its natural length, is this necessarily the amplitude that the mass oscillates through?

Answer 1

Solution : when a block is hung to a spring the block and spring has a new equilibrium position which can be calculated as

Kx = Mg here k is spring constant and x is extension from the natural length and m is mass of the block

So the length is extended from natural length as

X = mg / k

The spring acquires new MP here. As spring block was released suddenly so the spring would oscillate about its mean position with an amplitude of x = mg / k

The spring would oscillate with this amplitude if no non conservative forces are present and spring is considered to be ideal. So the elongation is 1 meter at equilibrium then oscillation would be of 1 m amplitude. However In real life some variation is seen due to mass of spring and air resistance.