Final answer:
To find the positions where the speed of a particle in SHM is one fifth of its maximum speed, solve the equation (1/5) Vmax = ± Vmax √(1 - (x/X)^2). The positions are at ± 2.94 cm from the equilibrium, where the amplitude X is 3.00 cm.
Step-by-step explanation:
To determine at what positions a particle executing simple harmonic motion (SHM) has a speed equal to one fifth of its maximum speed, we need to use the relationship between velocity and position in SHM. The maximum speed Vmax occurs at the equilibrium position (x=0), and the velocity at any other position can be found using the equation:
v = ± Vmax √(1 - (x/X)^2)
where x is the position of the particle, and X is the amplitude of the motion. For the velocity v to be 1/5 of Vmax, the equation becomes:
(1/5) Vmax = ± Vmax √(1 - (x/X)^2)
Solving for x/X:
x/X = ± √(1 - (1/25))
x/X = ± √(24/25)
This simplifies to:
x/X = ± 0.98
Since the amplitude X is 3.00 cm, we then calculate:
x = ± 0.98 * 3.00 cm
x = ± 2.94 cm
So, the positions where the speed of the particle is one fifth of its maximum speed are ± 2.94 cm from the equilibrium position.