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A person normally weighing 60 kg stands ...

A person normally weighing `60 kg` stands on a platform which oscillates up and down harmonically at a frequency `2.0 sec^(-1)` and an amplitude `5.0 cm` . If a machine on the platform gives the person's weight against time deduce the maximum and minimum reading it will shown, `Take g = 10 m//sec^(2)`.

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To solve the problem, we need to determine the maximum and minimum readings of the weight of a person standing on an oscillating platform. The weight readings will vary due to the oscillation of the platform. ### Step-by-Step Solution: 1. **Identify the given values:** - Mass of the person, \( m = 60 \, \text{kg} \) - Frequency of oscillation, \( f = 2.0 \, \text{s}^{-1} \) - Amplitude of oscillation, \( A = 5.0 \, \text{cm} = 0.05 \, \text{m} \) ...
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