Question

A mass of 50.00g hangs from a 7.00cm-long spring that is rigidly attached to a ceiling. The mass is pulled down slightly, let go, and is observed to make 8 round trips (up and back down) in 14.00s. What is the stiffness constant for this spring

Answer 1

0.645 N/M

Step-by-step explanation:

Given

Mass=50.00g

We have to convert into the kg

So Mass =0.050 Kg

`Time\ = (14)/(8)\ = 1.75\ sec`

We know that

`T\ =2\ PI\sqrt{(M)/(K) }`........................Eq(1)

Where T= time

and M= Mass

K= Stiffness constant

On squaring both side we get

`K=(4\pi^(2) M)/(T^(2) )`....Eq(2)

Putting the value of M ,T and π in Eq(2) we get

K=0.645 N/M