A person of weight 500 N stands on a platform that oscillates with a frequency of 2 Hz. If the amplitude of oscillation is 5 cm, what will be its weight recorded by a machine on the platform at the lo

The weight recorded by the machine on the platform at the lowest point of oscillation can be calculated using the formula for simple harmonic motion:

Weight = mg + ma

Where:
m = mass of the person (weight/acceleration due to gravity)
g = acceleration due to gravity (9.81 m/s^2)
a = acceleration of the platform at the lowest point of oscillation

Given that the weight of the person is 500 N, we can calculate the mass of the person:

m = 500 N / 9.81 m/s^2
m ≈ 51 kg

The acceleration of the platform at the lowest point of oscillation can be calculated using the formula for simple harmonic motion:

a = -ω^2A

Where:
ω = angular frequency (2πf)
f = frequency of oscillation (2 Hz)
A = amplitude of oscillation (5 cm = 0.05 m)

ω = 2π(2 Hz) = 4π rad/s

a = -(4π)^2(0.05 m)
a ≈ -39.48 m/s^2

Now we can calculate the weight recorded by the machine on the platform at the lowest point of oscillation:

Weight = (51 kg)(9.81 m/s^2) + (51 kg)(-39.48 m/s^2)
Weight ≈ 500 N - 2000 N
Weight ≈ -1500 N

Therefore, the weight recorded by the machine on the platform at the lowest point of oscillation is approximately -1500 N. This negative value indicates that the person experiences a reduced weight due to the acceleration of the platform at that point.

At the lowest point of oscillation, the weight recorded by the machine will be the sum of the person's weight (500 N) and the maximum value of the centrifugal force.

Centrifugal force, Fc = mω²r

where:
m = mass of the person (500 N / 9.81 m/s²) = 51.02 kg
ω = angular frequency (2πf) = 2π(2 Hz) = 4π rad/s
r = amplitude of oscillation = 5 cm = 0.05 m

Fc = (51.02 kg)(4π rad/s)²(0.05 m) = 1005.3 N

Therefore, the weight recorded by the machine at the lowest point of oscillation will be:

Weight = Person's weight + Centrifugal force
Weight = 500 N + 1005.3 N
Weight = 1505.3 N