Question

MATHPHYS HELP
The force required to stretch a spring varies
directly with the amount the spring is
stretched. A spring stretches by 98 m when
a 78 N weight is hung from it and the weight
is at rest (at equilibrium). The 78 N weight
is replaced by an unknown weight W so that
the spring is stetched to a new equilibrium
position, 67 m below the position if no weight
were attached. The weight W is then dis-
placed from equilibrium and released so that
it oscillates.

Answer 1

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s