Question

Suppose you have three springs with force constants of k1 = k2 = k3 = 3.70 x 10^3 N/m. What is their effective force constant if one is hung from the other in series? You may assume the springs have negligible mass.

The answer should be in N/m.

Answer 1

When springs are connected in series, the effective force constant is obtained by summing the reciprocals of the individual force constants. In this case, since all three springs have the same force constant, the effective force constant will be the sum of the reciprocals of the force constants of the individual springs. Therefore, the effective force constant of the three springs when hung in series is 1.23 x 10^3 N/m.

Step-by-step explanation:

When springs are connected in series, the effective force constant is obtained by summing the reciprocals of the individual force constants. In this case, since all three springs have the same force constant, the effective force constant will be the sum of the reciprocals of the force constants of the individual springs.

The force constant of each spring is k = 3.70 x 10^3 N/m. Therefore, the effective force constant when the springs are hung in series is:

1/keff = 1/k1 + 1/k2 + 1/k3

1/keff = 1/(3.70 x 10^3 N/m) + 1/(3.70 x 10^3 N/m) + 1/(3.70 x 10^3 N/m)

1/keff = 3/(3.70 x 10^3 N/m)

keff = (3.70 x 10^3 N/m)/3 = 1.23 x 10^3 N/m

Therefore, the effective force constant of the three springs when hung in series is 1.23 x 10^3 N/m.

Answer 2

The effective force constant is 1233.33 N/m.

Step-by-step explanation:

It is given that,

Force constant 1,
`k_1=3.7* 10^3\ N/m`

Force constant 2,
`k_2=3.7* 10^3\ N/m`

Force constant 3,
`k_3=3.7* 10^3\ N/m`

The effective force constant if one is hung from the other in series is given by :

`(1)/(K_(eff))=(1)/(k_1)+(1)/(k_2)+(1)/(k_3)`

`(1)/(K_(eff))=(1)/(3.7* 10^3)+(1)/(3.7* 10^3)+(1)/(3.7* 10^3)`

`K_(eff)=1233.33\ N/m`

So, the effective force constant is 1233.33 N/m. Hence, this is the required solution.