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.
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
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1 answer