Answer:
a
The new frequency is 2.0 Hz
b
The new amplitude is [tex]A_1 = 0.0120 \ m = 1.2 \ cm[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m = 200 = 0.20 \ kg[/tex]
The first amplitude is [tex]A = 2.0 \ cm = 0.02 \ m[/tex]
The frequency is [tex]f = 2.0 \ Hz[/tex]
The force exerted is F = 20 N
The duration of the force is [tex]t = 1.0 ms = 1.0*10^{-3} \ s[/tex]
Generally the impulse is mathematically represented as
[tex]I = F * t[/tex]
=> [tex]I = 20 * 1.0 *10^{-3}[/tex]
=> [tex]I = 0.02 \ kg \ m/s[/tex]
Generally the initial angular speed of the block is mathematically represented as
[tex]w_1 = 2\pi f[/tex]
=> [tex]w_1 = 2 * 3.142 * 2[/tex]
=> [tex]w_1 = 12.568 \ rad/s[/tex]
Generally the linear velocity of the block at the equilibrium position before the impact is mathematically represented as
[tex]v_1 = A * w_1[/tex]
=> [tex]v_1 =0.02 *12.568[/tex]
=> [tex]v_1 =0.2513 \ m/s[/tex]
Generally the change in velocity after that impact of the force is mathematically represented as
[tex]\delta v = \frac{I}{m}[/tex]
=> [tex]\delta v = \frac{0.02}{0.20 }[/tex]
=> [tex]\delta v = 0.1[/tex]
Generally the linear velocity of the block at the equilibrium position after the impact is mathematically represented as
[tex]v_2 = v_1 - \delta v[/tex]
=> [tex]v_2 = 0.2513 - 0.1[/tex]
=> [tex]v_2 = 0.1513[/tex]
This can also be mathematically represented as
[tex]v_2 = A_1 * w[/tex]
=> [tex]0.1513 = A_1 * 12.568[/tex]
=>[tex]A_1 = 0.0120 \ m = 1.2 \ cm[/tex]
Generally given that the angular velocity does not change , it then implies that the frequency remains 2.0 Hz
