Question

A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?

Answer 1

2

Step-by-step explanation:

pulling force because of it force

Answer 2

5.9 cm

Step-by-step explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

`f = (1)/(2\pi) \sqrt{ (k)/(m) }`

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

`k = {(2\pi \: f )}^(2) m = {(2\pi(1.4hz))}^(2) 0.26kg = 20.1 (n)/(m)`

The restoring force of the spring is:

F= kx

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

`x = (f)/(k) = (1.2 \: n)/(20.1 (n)/(m) ) = 0.059m \: = 5.9 \: cm`