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?

From the question we are told that

The mass of the block is $m=200g=0.2kg$

The first amplitude is $A=2cm=0.02m$

The frequency is $f=2Hz$

The force exerted is $F=20N$

The duration of the force is $t=1ms=1\times {10}^{-3}s$

Generally the impulse is mathematically represented as

$I=F\times t$

$I=20\times 1\times {10}^{-3}$

$I=0.02kgm/s$

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

localid="1650202224531" $$\begin{gathered} {\omega }_1=2\mathrm{\pi f} \\ {\mathrm{\omega }}_1=2\times 3.142\times 2 \\ {\mathrm{\omega }}_1=12.568\mathrm{rad}/\sec \end{gathered}$$

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

$$\begin{gathered} v_1=A{\omega }_1 \\ {v}_1=0.02\times 12.568 \\ {v}_1=0.2513m/s \end{gathered}$$

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

$$\begin{gathered} \delta v=\frac{I}{m} \\ \delta v=\frac{0.02}{0.20} \\ \delta v=0.1 \end{gathered}$$

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

$$\begin{gathered} v_2=v_1-\delta v \\ {v}_2=0.2513-0.1 \\ {v}_2=0.1513 \end{gathered}$$

This can also be mathematically represented as ,

$$\begin{gathered} v_2=A_1\omega \\ 0.1513=A_1\times 12.568 \\ {A}_1=0.012m \\ =1.2cm \end{gathered}$$

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