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GarryVolchara [31]
3 years ago
10

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.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

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{\frac{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= \frac{V_{max}}{w}

Computing w

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

Calculating A

\displaystyle A=\frac{1.7}{24.5}=0.069\ m

\displaystyle \boxed{A=6.9\ cm}

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