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NCERT Exemplar · Q35

Q.A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s−12.0\ \mathrm{s^{-1}} and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part
(a) is yes, what will be the maximum and minimum reading in the machine and at which position?
Uttarakhand UbseLong· 3mImportance★★★★★est
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A weighing machine reads the normal force, which changes with the platform's acceleration in SHM. With amax=ω2A≈7.9 m/s2a_\text{max}=\omega^2A\approx7.9\ \text{m/s}^2, the reading is maximum ≈90.3\approx 90.3 kg at the lowest point and minimum ≈9.7\approx 9.7 kg at the highest point.

(a) Does the reading change?

Yes. The machine measures the normal force NN it exerts on the person, not the true weight mgmg. In SHM the platform accelerates as a=−ω2xa=-\omega^2x, so NN must supply the net force mama and therefore varies through the cycle.

(b) Maximum and minimum readings

  1. Angular frequency. With f=2.0 s−1f = 2.0\ \text{s}^{-1},

ω=2πf=4π rad/s,ω2=16π2≈158 rad2/s2.\omega = 2\pi f = 4\pi\ \text{rad/s},\qquad \omega^2 = 16\pi^2 \approx 158\ \text{rad}^2/\text{s}^2.

  1. Maximum acceleration. With amplitude A=5.0 cm=0.05 mA = 5.0\ \text{cm} = 0.05\ \text{m},

amax=ω2A=158×0.05≈7.9 m/s2.a_\text{max} = \omega^2 A = 158\times0.05 \approx 7.9\ \text{m/s}^2.

  1. Newton's second law (upward positive): N−mg=maN - mg = ma, so N=m(g+a)N = m(g+a). The machine reading in kilograms is N/gN/g.

  2. Lowest point — platform accelerates upward, a=+amaxa=+a_\text{max}: …

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