Question

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximum speed?​

Answer 1

2.34

Step-by-step explanation:

To find the position where the speed is half of its maximum speed, we can set v = v_max/2 and solve for x:

v_max/2 = ω√(A^2 - x^2)

Substituting the given values, we have:

(ωA)/2 = ω√(A^2 - x^2)

Simplifying and rearranging:

A^2 - x^2 = (A/2)^2

x^2 = A^2 - (A/2)^2

x^2 = (3.00 cm)^2 - (1.50 cm)^2

x = √(6.75 cm^2 - 2.25 cm^2)

x = √5.50 cm^2

x ≈ 2.34 cm

Therefore, the position where the speed of the particle is half of its maximum speed is approximately 2.34 cm from the equilibrium position.