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 :

elcharly64

Answer:

A = 6.9 cm

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

[tex]\displaystyle w=\sqrt{\frac{k}{m}}[/tex]

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

[tex]\displaystyle X=A\ cos\ w\ t[/tex]

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

[tex]\displaystyle X'=V=-A\ w\ sin\ w\ t[/tex]

And the maximum speed is

[tex]\displaystyle V_{max}= A\ w[/tex]

Solving for the amplitude

[tex]\displaystyle A= \frac{V_{max}}{w}[/tex]

Computing w

[tex]\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s[/tex]

Calculating A

[tex]\displaystyle A=\frac{1.7}{24.5}=0.069\ m[/tex]

[tex]\displaystyle \boxed{A=6.9\ cm}[/tex]

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