Question

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 1.7 m/s. Determine the amplitude A of the motion.

Answer 1

A = 6.9 cm

Step-by-step explanation:

Simple Harmonic Motion

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

`\displaystyle w=\sqrt{(k)/(m)}`

The equation for the motion of the object is written as a sinusoid:

`\displaystyle X=A\ cos\ w\ t`

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

`\displaystyle X'=V=-A\ w\ sin\ w\ t`

And the maximum speed is

`\displaystyle V_(max)= A\ w`

Solving for the amplitude

`\displaystyle A= (V_(max))/(w)`

Computing w

`\displaystyle w =\sqrt{(120)/(0.2)}=24.5\ rad/ s`

Calculating A

`\displaystyle A=(1.7)/(24.5)=0.069\ m`

`\displaystyle \boxed{A=6.9\ cm}`