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

a

The new frequency is 2.0 Hz

b

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

Step-by-step 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 = (I)/(m)`

=>
`\delta v = (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