Question

A 200 g block attached to a horizontal spring is oscillating with an amplitude of 2.0 cm and a frequency of 2.0 Hz. Just as it passes through the equilibrium point, moving to the right, a sharp blow directed to the left exerts a 20 N force for 1.0 ms. What are the new (a) frequency and (b) amplitude

Answer 1

Answer:

a

The new frequency is  2.0 Hz

b

The new amplitude is $A_1 = 0.0120 \ m = 1.2 \ cm$

Explanation:

From the question we are told that

    The mass of the block is  $m = 200 = 0.20 \ kg$

     The first  amplitude is  $A = 2.0 \ cm = 0.02 \ m$

      The frequency is $f = 2.0 \ Hz$

      The force exerted is  F = 20 N

      The duration of the force is  $t = 1.0 ms = 1.0*10^{-3} \ s$

Generally the impulse is mathematically represented as

         $I = F * t$

=>      $I = 20 * 1.0 *10^{-3}$

=>      $I = 0.02 \ kg \ m/s$

Generally the initial angular speed of the block is mathematically represented as

       $w_1 = 2\pi f$

=>    $w_1 = 2 * 3.142 * 2$

=>    $w_1 = 12.568 \ rad/s$

Generally the  linear velocity of the block at the equilibrium position before the impact  is mathematically represented as

       $v_1 = A * w_1$

=>    $v_1 =0.02 *12.568$

=>  $v_1 =0.2513 \ m/s$

Generally the change in velocity after that impact of the force is mathematically represented as

        $\delta v = \frac{I}{m}$

=>     $\delta v = \frac{0.02}{0.20 }$

=>     $\delta v = 0.1$

Generally the  linear velocity of the block at the equilibrium position after the impact  is mathematically represented as  

        $v_2 = v_1 - \delta v$

=>    $v_2 = 0.2513 - 0.1$

=>    $v_2 = 0.1513$

This can also be mathematically represented as

     $v_2 = A_1 * w$

=>   $0.1513 = A_1 * 12.568$

=>$A_1 = 0.0120 \ m = 1.2 \ cm$

Generally given that the angular velocity does not change , it then implies that the frequency remains 2.0 Hz