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

Answer:

The value is  $x = 0.084 \ m$

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

      $k = \frac{ F }{x }$

=>   $14.33 = \frac{ 1.2 }{x }$

=>   $x = 0.084 \ m$