A particle executes simple harmonic motion with an amplitude of 9.00 cm. At what positions does its speed equal two thirds of its maximum speed?

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To find the positions at which the speed of the particle equals two thirds of its maximum speed, we first need to understand the relationship between speed and position in simple harmonic motion.

In simple harmonic motion, the speed of the particle is maximum at the equilibrium position (the center of the motion) and decreases as the particle moves away from the equilibrium position. The speed is zero at the turning points, where the particle changes direction.

The equation that relates the speed (v) of the particle to its position (x) in simple harmonic motion is:

v = ω √(A^2 - x^2)

Where:
v = speed of the particle
ω = angular frequency of the motion
A = amplitude of the motion
x = position of the particle

In this case, we know the amplitude of the motion (A) is 9.00 cm. We also know that the speed is two-thirds of its maximum speed. Let's denote the maximum speed as vmax and the speed at two-thirds of the maximum speed as v_2/3.

v_2/3 = (2/3) * vmax

To find the positions at which the speed equals two-thirds of the maximum speed, we need to solve for x in the equation above.

To do this, we can rearrange the equation as follows:

(2/3) * vmax = ω √(A^2 - x^2)

Now, we substitute the values we know:
A = 9.00 cm
v_2/3 = (2/3) * vmax

Since we don't know the value of vmax or ω, we can't directly find x. However, we can make a useful observation. The equation for the speed (v) in simple harmonic motion shows that the speed is zero at the turning points (x = ±A). Therefore, we can conclude that the speed will be equal to two-thirds of the maximum speed at positions where the particle is a distance A/sqrt(3) away from the equilibrium position.

So, the positions where the speed equals two-thirds of the maximum speed are located at:

x = A/sqrt(3) = 9.00 cm / sqrt(3) ≈ 5.20 cm

Therefore, the particle's speed will be two-thirds of its maximum speed at positions approximately 5.20 cm away from the equilibrium position in either direction.