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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)`.

Text Solution

Verified by Experts

The correct Answer is:
`107.4`
The platform vibrates between the positions A and B about the mean position Q as shown in

Given A = 5 . 0 cm
` m = 60 kg, v = 2 Hz `
At A and B the acceleration is maximum and is directed towards the mean position .
It is given by
` a_("max") = omega ^(2) A `
`= 4 pi ^(2) v^(2) A `
`= 4 xx 9 . 87 xx (2) ^(2) xx 0.05 = 7.9 ms^(-2)`
At A, both the weight mg and the restoring force F and directed towards O .Therefore , the weight at A is maximum and is given by
`W_(1) = (mg + F) = (mg + ma_("max") ) = m ( g + a_("max"))`
` = 60 ( 10 + 7.9) = 60 xx 17.9 = 1074 N`
`= (1074)/(g) = (1074)/(10) = 107.4 kg f `
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