Question

a particle is moving in simple harmonic motion with an amplitude of 5.0 cm and a maximum velocity of 15.0 cm/s. assume the initial phase is 0. (include the sign of the value in your answer.) (a) at what position is its velocity 4.3 cm/s? cm (b) what is its velocity when its position is 2.5 cm? cm/s

Answer 1

The position at which the velocity is 4.3 cm/s is approximately 4.71 cm. The velocity when the position is 2.5 cm is approximately 11.59 cm/s.

(a) To find the position at which the velocity is 4.3 cm/s, we can use the equation for velocity in simple harmonic motion: v = ω√(A² - x²). Rearranging the equation to solve for x, we have: x = √(A² - (v/ω)²). Plugging in the given values, we get: x = √(5.0² - (4.3/15.0)²) ≈ 4.71 cm. The position at which the velocity is 4.3 cm/s is approximately 4.71 cm.

(b) To find the velocity when the position is 2.5 cm, we can use the equation for velocity in simple harmonic motion: v = ω√(A² - x²). Rearranging the equation to solve for v, we have: v = ω√(A² - x²). Plugging in the given values, we get: v = 15.0√(5.0² - 2.5²) ≈ 11.59 cm/s. The velocity when the position is 2.5 cm is approximately 11.59 cm/s.