Final answer:
To find the positions where the particle's speed is two-thirds of the maximum in SHM, we use the energy conservation principle and calculate that the positions are approximately +1.49 cm and -1.49 cm from the equilibrium.
Step-by-step explanation:
The question is asking about a particle executing simple harmonic motion (SHM) and we need to find the positions at which its speed equals two-thirds of its maximum speed. The maximum speed is attained when the particle passes through the equilibrium position (x=0). We can use the energy conservation principle, where the sum of kinetic and potential energy is constant at any point in SHM.
Let's denote the maximum speed as Vmax. According to the conservation of energy in SHM, we have:
(1/2)mv^2 + (1/2)kx^2 = (1/2)kA^2, where m is mass, v is the speed of the particle at position x, k is the spring constant, x is the displacement from equilibrium, and A is the amplitude. For the speed to be two-thirds of the maximum speed, we have v = (2/3)Vmax.
Plugging v = (2/3)Vmax into the energy equation and solving for x, we get the following equation:
(1/2)m((2/3)Vmax)^2 + (1/2)kx^2 = (1/2)kA^2. Simplifying this gives the ratio of displacement to amplitude as x/A = ±\sqrt{1 - (4/9)}. Calculating this provides us with x/A = ±\sqrt{5/9} = ±\sqrt{5}/3.
Substituting the given amplitude A = 2.00 cm into the equation, we find the positions as ±(\sqrt{5}/3) × 2.00 cm.
SUM UP all the final answer as points at last: The positions where the speed is two-thirds of the maximum speed are approximately:
- Larger number: +1.49 cm
- Smaller number: -1.49 cm