Question

A 0.50-kg box is attached to an ideal spring of force constant (spring constant) 20 N/m on a horizontal, frictionless floor. The box oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) What is the amplitude of vibration

Answer 1

`A = 0.24 m`

Step-by-step explanation:

As we know that for spring block system the angular frequency is given as

`\omega = \sqrt{(k)/(m)}`

now we know that

`k = 20 N/m`

also we know that

`m = 0.50 kg`

so we have

`\omega = \sqrt{(20)/(0.5)}`

`\omega = 6.32 rad/s`

now we know that speed of an SHM at its equilibrium position is given as

`v = A\omega`

now plug in all values in it

`1.5 = A(6.32)`

`A = 0.24 m`