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in Physics by (69.2k points)

A person normally weighing 60 kg stands on a plat form which oscillates up and down harmonically at a frequency 2.0 sec –1 and an amplitude 5.0 cm. If a machine on the plat form gives the person’s weight against time, deduce the maximum and minimum reading it will show, take g = 10 m/sec 2

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As shown in figure, plat form is executing SHM with amplitude and angular frequency give

A = 5.0 cm

and ω = 2πn = 4π rad/sec [as n = 2 sec –1 ]

Here weighing machine will show weight more then that of man when it is below its equailibrium position when the acceleration of platform is in upward direction. In this situation the free body diagram of man relative to plat form is shown in figure. Here ma is a pseudo force on man in downward direction relative to platform(or weighing machine).As weighing machine will read the normal reaction on it thus for equilibrium of an relative to platform,

we have N = mg + ma

or N = mg + m(ω2 y) [as | a | = ω2 y]

Where y is the displacement of plat form from its mean position. We wish to find the maximum weight shown by the weigh in gm a chine, which is possible when plat form is at its lowest extreme position as shown in figure, thus maximum reading of weigh in gm a chine will be

N = mg + mω2A

N = 60 × 10 + 60 × (4π)2 × 0.05

N = 600 + 480

N = 1080 Nt = 108 kg ωt

Similarly the machine will show minimum reading when it is at it supper extreme position when pseudo force on man will be in upward direction, thus minimum reading of weighing machine will be

N = mg – mω2A

N = 600 – 480 = 120 Nt = 12 kg ωt

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