an object with mass 2.0 kgkg is executing simple harmonic motion, attached to a spring with spring constant 280 n/mn/m . when the object is 0.020 mm from its equilibrium position, it is moving with a speed of 0.40 m/sm/s . a) Calculate the amplitude of the motion.b) Calculate the maximum speed attained by the object.

Respuesta :

The amplitude is ≈ 10.18 s-1.

What is amplitude ?

The greatest displacement or distance that a point on a wave or vibrating body can travel in relation to its equilibrium position It is equivalent to the length of the vibration path divided in half.

What is speed ?

The speed at which an object's location changes in any direction. The distance traveled in relation to the time it took to travel that distance is how speed is defined.

Speed = v = 0.55 m/s , given

Angular speed = ω , where:

W= [tex]\sqrt{K/M}[/tex]= [tex]\sqrt{280 N/M/2.7Kg}[/tex]= 10.18350154 =[tex]\sqrt{kg*m/ kg*s}[/tex]

≈ 10.18 s-1

Amplitude = A, where:

A =[tex]\sqrt{x0+v0/w2}[/tex]

Therefore, the amplitude is ≈ 10.18 s-1.

Learn more about amplitude from the given link.

https://brainly.com/question/21632362

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