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?

I got:

v2 +ω^2x^2 =ω^2A^2

vmax =ω A and v = ωA/2?

Since it executes SHM, the sum of potential energy and kinetic energy is a constant, with PE proportional to x^2 and KE proportional to V^2. This means

(x/A)^2 + (V/Vmax)^2 = 1

where A is the amplitude and x is the distance from the mean position.

When V/Vmax = 2/3,
(x/A)^2 = 1 - 4/9 = 5/9
x/A = (sqrt5)/3 = (+or-) 0.7454

To find the positions where the speed of the particle equals two-thirds of its maximum speed, we can use the equation v^2 + ω^2x^2 = ω^2A^2, where v is the speed of the particle, ω is the angular frequency, x is the displacement from the equilibrium position, and A is the amplitude.

For simple harmonic motion, the maximum speed (vmax) is equal to ωA.

Since we are looking for a speed that is two-thirds of the maximum speed (vmax), we can write:

v = (2/3)vmax

Substituting vmax = ωA into the equation, we get:

(2/3)vmax = (2/3)ωA

Simplifying, we find:

v = (2/3)ωA

Now let's substitute this into the equation v^2 + ω^2x^2 = ω^2A^2:

[(2/3)ωA]^2 + ω^2x^2 = ω^2A^2

(4/9)(ω^2A^2) + ω^2x^2 = ω^2A^2

Multiplying through by 9/4, we get:

(9/4)(4/9)(ω^2A^2) + (9/4)ω^2x^2 = (9/4)ω^2A^2

Simplifying, we find:

ω^2x^2 = (9/4)ω^2A^2 - (9/4)(4/9)(ω^2A^2)

ω^2x^2 = (9/4)(ω^2A^2 - ω^2A^2/9)

ω^2x^2 = (9/4)(8/9)(ω^2A^2)

ω^2x^2 = (2/1)(1/1)(ω^2A^2)

ω^2x^2 = 2ω^2A^2

x^2 = 2A^2

Taking the square root of both sides, we find:

x = ±√2A

Since the amplitude A = 9.00 cm, the positions where the speed of the particle equals two-thirds of its maximum speed are:

x = ±√2(9.00 cm)

x ≈ ±12.73 cm

Therefore, the positions where the speed of the particle equals two-thirds of its maximum speed are approximately 12.73 cm and -12.73 cm.

To find the positions where the speed of the particle is equal to two-thirds of its maximum speed, we can use the formula for velocity in simple harmonic motion.

The formula for velocity in simple harmonic motion is given by:
v = ω√(A^2 - x^2)

Where:
v is the velocity of the particle,
ω is the angular frequency (2πf),
A is the amplitude of the motion, and
x is the position of the particle.

In this case, we want to find the positions where the speed (v) is equal to two-thirds of the maximum speed (vmax).

We know that the maximum speed (vmax) is given by:
vmax = ωA

Substituting this value into the velocity formula, we can rewrite it as:
v = (ωA)√(1 - (x^2/A^2))

Now we can set v equal to two-thirds of vmax and solve for x:
(2/3)vmax = (2/3)ωA = ωA/3

(ωA/3) = (ωA)√(1 - (x^2/A^2))

Dividing both sides by ωA:
1/3 = √(1 - (x^2/A^2))

Squaring both sides to eliminate the square root:
1/9 = 1 - (x^2/A^2)

Rearranging the equation:
x^2/A^2 = 1 - 1/9

Simplifying:
x^2/A^2 = 8/9

Cross-multiplying:
x^2 = 8A^2/9

Taking the square root:
x = ± √(8A^2/9)

Since the amplitude (A) is given as 9.00 cm, we can substitute the value and find the positions where the speed is two-thirds of the maximum speed:
x = ± √(8 * 9.00^2 / 9) = ± √(8 * 81 / 9) = ± √(8 * 9) = ± √(72) = ± 2√(18) ≈ ± 2√(9) ≈ ± 2 * 3 ≈ ± 6.

Therefore, the positions where the speed of the particle is equal to two-thirds of its maximum speed are approximately ±6 cm from the equilibrium position.