The given motion equation is:

$x(t) = 3 \sin(20\pi t)$

This represents a sinusoidal oscillation, where $x(t)$ is the position of the object along the x-axis in centimeters, and $t$ is time in seconds. The general form for sinusoidal motion is:

$x(t) = A \sin(\omega t)$

where:

• $A$ is the amplitude of oscillation (in this case, $A = 3$ cm),
• $\omega$ is the angular frequency (in this case, $\omega = 20\pi$).

Step 1: Calculate the frequency of oscillation

The angular frequency $\omega$ is related to the frequency $f$ by the formula:

$\omega = 2\pi f$

Substituting $\omega = 20\pi$:

$20\pi = 2\pi f$

Solving for $f$:

$f = \frac{20\pi}{2\pi} = 10 \, \text{Hz}$

This means the object completes 10 oscillations per second.

Step 2: Back-and-forth motion

One complete back-and-forth motion (from the origin, to the right, back to the origin, to the left, and back to the origin) corresponds to one full cycle of the sine function. Therefore, the object makes 10 complete back-and-forth motions in one second.

Final Answer:

The object makes 10 complete back-and-forth motions in one second.

Would you like further details or have any questions?

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Relative Questions:

1. How would the amplitude affect the motion in this case?
2. What is the period of the object's motion?
3. How would the motion change if the angular frequency were halved?
4. What is the maximum velocity of the object?
5. How would the motion change if the equation were $x(t) = 3 \cos(20\pi t)$?

Tip: The frequency of a sinusoidal motion can be easily determined from the angular frequency using the formula $\omega = 2\pi f$.