Question

The force required to stretch a spring varies directly with the amount the spring is stretched. A spring stretches by 33 m when a 19 N weight is hung from it and the weight is at rest (at equilibrium). The 19 N weight is replaced by an unknown weight W so that the spring is stetched to a new equilibrium position, 17 m below the position if no weight were attached. The weight W is then displaced from equilibrium and released so that it oscillates.

Answer 1

Answer:

The period is  $T = 8.27 \ sec$

Explanation:

From the question we are told that

   The extension of the spring is  $e_1 = 33 \ m$

   The  first weight  applied is  $F_1 = 19 N$

     The second weight applied is  $F_2 = W$

     The second extension is $e_2 = 17 \ m$

The spring constant of the spring is mathematically evaluated as

         $k = \frac{F_1}{e_1 }$

substituting values

       $k = \frac{19}{33 }$

       $k = 0.576$

We are told that

         19 N extended the spring to 33 m    

Then W N  will extended it by  17 m

Therefore     $W = \frac{19 * 17}{33}$

                    $W = 9.788 \ N$

Generally the period of the oscillation is mathematically represented as

           $T = 2 \pi \sqrt{\frac{M}{K} }$

where M is the mass of the W which is mathematically evaluated as

         $M = \frac{9.788}{9.8}$

          $M = 1.0 \ kg$

substituting values

        $T = 2 \pi \sqrt{\frac{1}{0.576} }$

       $T = 8.27 \ sec$