The period of oscillation of a particle executing or undergoing simple harmonic motion is 45 sec if the amplitude of the oscillation is 3m calculate the maximum speed of the particle
The period of oscillation (T) is the time it takes for one complete cycle of oscillation. In this case, T = 45 sec.
The formula for the period of oscillation is given by:
T = 2π√(m/k)
Where m is the mass of the particle and k is the spring constant.
In simple harmonic motion, the formula for the maximum speed (vmax) can be given as:
vmax = Aω
Where A is the amplitude of oscillation and ω is the angular frequency.
From the period formula, we can rearrange it to solve for ω:
ω = 2π/T
With A = 3m and ω = 2π/T, we can calculate the maximum speed:
vmax = 3m * 2π/T
Substituting T = 45 sec and simplifying:
vmax = 3m * 2π/45 sec
Calculating the value:
vmax ≈ 0.418 m/s
Therefore, the maximum speed of the particle is approximately 0.418 m/s.