Question

A man of mass 60kg standing on a platform executing
SHM in a vertical plane. The displacement from mean
position is y= 0.5 sin (2 teft). The minimum value of
f for which the man will feel weightlessness at the
highest point is​

Answer 1

The minimum value of f is $\dfrac{\sqrt{2g}}{2\pi}$

Explanation:

Given that,

Mass of man = 60 kg

The displacement equation from mean position

$y=0.5\sin(2\pi f t)$

We need to calculate the minimum value of f

Using formula of maximum acceleration

$\omega^2 A=g$

$(2\pi f)^2A=g$

$f^2=\dfrac{g\times 2}{4\pi^2}$

$f=\dfrac{\sqrt{2g}}{2\pi}$

Hence, The minimum value of f is $\dfrac{\sqrt{2g}}{2\pi}$

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Topic : minimum frequency

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