could you please show me how to do this?

Question

A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period
of 2.00 s. When the 0.750-kg object is removed and replaced by a 1.25-kg object, what
will be the period of oscillation?

Answer 1

hmm I assume this all occurs on Earth at altitude of 0m?
Well then gravity equals g=9.81 m/s^2 at 0m altitude on earth
so the force of the object = 0.750kg*9.81m/s^2
F = ma = 0.750kg*(gravitational acceleration)
P = 1/f Period equals the inverse of frequency
so f = 1/P = 1/2s = 0.5 Hertz = 0.5 (1/s)
so distance equals (2s)^2 * 9.81 m/s^2 = 39.24 m
F = kx (Hooke's law)
kx = mg because the spring doesnt snap, the forces of gravitation is equal to the spring force in opposite directions.
k is the spring constant and since both masses are using the same spring this can be ignored.
k*39.24 = 7.3575m/s^2 = mg
k = 0.1875
now we have:
k*x = mg for the second mass on the same spring
0.1875*x = 1.25 kg* 9.81m/s^2
x = 65.4m (second mass)
x/9.81m/s^2 = 65.4m/g = 6.6666
P = 20/3 s

which is totally incorrect!

2
2 4 * pi
T = ------- * m
k
this is the equation for Period...hmmm the same spring means the same spring constant so the period is solely dependent upon the spring constant and not the mass!
so Period = 2s, this is because the object travels a greater distance but actually moves at a faster velocity so the period and frequency remain unchanged as long as the spring remains the same!


Why did I show all that work above? To show you what students normally do when equations are better at showing relationships.

Answer 2

that wil be exactly 3.3333333333 second try solving it by proportion