A massless spring having a spring constant k = 8.75 N/m is hung vertically. If t

Question

A massless spring having a spring constant k = 8.75 N/m is hung vertically. If the spring is displaced 0.150 m from its equilibrium position, what is the force that the spring exerts? A 400-g mass is suspended from this spring. What is the displacement of the end of the spring due to the weight of the mass? Suppose this mass is allowed to oscillate on the spring. What is the period of the oscillation? What is the frequency of the oscillation? A 0.100-kg mass suspended vertically on a spring takes 10.94 s to undergo 20 oscillations. What is the spring constant of the spring?

Explanation / Answer

1. whether the displacement is left or right, and whether left or right is positive,

I would definitely stick with F = kx where

k= spring constant
F=force
x=extention of spring

so, F = 8.75 X0.150 = 1.3125 N

2.)

F = kx = mg

x = mg / k = (0.5)(9.8) / (8.75) = 0.56 m = 56 cm

so, the displacement of the spring is 56 cm

3) so, period of oscillation = 2pi sqrt(x/g) = 2x3.14 sqrt(0.56/9.8) = 1.50 seconds

4) frequency of oscillation = 1/T = 1/1.50 = 0.6667 Hz

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