Question

A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should the block's mass be in order to reduce the frequency to half its initial value? Express your answer in terms of the variables m and A.

Answer 1

Answer:

The block's mass should be $3m$

Explanation:

Given:

Cart with mass $m$

From the conservation of energy before mass is added,

  $\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}$

Where $A =$ amplitude of spring mass system, $k =$ spring constant

  $A = v\sqrt{\frac{m}{k} }$

Now new mass $M$ is added to the system,

   $\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}$

  $A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

   $f' = \frac{f}{2}$

Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

        $\omega ' = \frac{\omega}{2}$

$\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}$

   $\frac{k}{m +M } = \frac{k}{4m}$

   $M = 3m$

Therefore, the block's mass should be $3m$