A horizontal platform with an object placed on it is executing SHMSHM in the vertical direction. The amplitude of oscillation is 2.5 cm2.5 cm . What must be the least period of these oscillation so that the object is not detached ?
Question
A horizontal platform with an object placed on it is executing SHMSHM in the vertical direction. The amplitude of oscillation is 2.5 cm2.5 cm . What must be the least period of these oscillation so that the object is not detached ?
Solution
The object will detach from the platform when the acceleration of the platform is equal to or greater than the acceleration due to gravity (g).
The maximum acceleration (a_max) of an object in Simple Harmonic Motion (SHM) is given by the formula:
a_max = (2π/T)^2 * A
where:
- T is the period of the oscillation,
- A is the amplitude of the oscillation.
We know that a_max should be equal to g for the object to just about to detach. So, we can set up the equation:
g = (2π/T)^2 * A
We know that g = 9.8 m/s^2 and A = 2.5 cm = 0.025 m. We can substitute these values into the equation and solve for T:
9.8 = (2π/T)^2 * 0.025
Rearranging the equation to solve for T gives:
T = 2π * sqrt(A / g)
Substituting the known values gives:
T = 2π * sqrt(0.025 / 9.8)
Solving this equation gives:
T ≈ 0.318 s
So, the least period of the oscillation for the object to not detach from the platform is approximately 0.318 seconds.
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