Question

A block with a mass of 0.30 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 1.1 Hz.

How far was the block pulled back before being released?

Answer 1

The value is
`x = 0.084 \ m`

Step-by-step explanation:

From the question we are told that

The mass of the block is
`m = 0.30 \ kg`

The force exerted is
`F = 1.2 \ N`

The frequency is
`f = 1.1 \ Hz`

Generally the spring constant of the spring is mathematically represented as

`k = (2 \pi f)^2 * m`

=>
`k = (2 * 3.142 * 1.1 )^2 * 0.30`

=>
`k = 14.33 \ N/m`

Generally the spring constant is also mathematically represented as

`k = ( F )/(x )`

=>
`14.33 = ( 1.2 )/(x )`

=>
`x = 0.084 \ m`