A person normally weighing 50 kg50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 3.0 s−13.0 s-1 and an amplitude 6.0 cm6.0 cm. A weighing machine on the platform gives the person's weight against time. The maximum reading in the machine is X𝑋units. Then the value of X10𝑋10 must have been _____ (Take π2=10𝜋2=10 )
Question
A person normally weighing 50 kg50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 3.0 s−13.0 s-1 and an amplitude 6.0 cm6.0 cm. A weighing machine on the platform gives the person's weight against time. The maximum reading in the machine is X𝑋units. Then the value of X10𝑋10 must have been _____ (Take π2=10𝜋2=10 )
Solution
The weight of a person in an oscillating lift varies between mg(1 - (2πf)²) and mg(1 + (2πf)²), where m is the mass of the person, g is the acceleration due to gravity, and f is the frequency of oscillation.
Given that the mass m = 50 kg, g = 10 m/s² (approx), and f = 3.0 s⁻¹, we can substitute these values into the equation for the maximum weight:
W_max = mg(1 + (2πf)²) = 50 kg * 10 m/s² * (1 + (2π3.0 s⁻¹)²) = 500 kgm/s² * (1 + 36π²) = 500 kgm/s² * (1 + 360) (since π² is approximated as 10) = 500 kgm/s² * 361 = 180500 kgm/s² = 180500 N (since 1 kg*m/s² = 1 N)
Therefore, the maximum reading on the weighing machine is 180500 N. If the machine measures weight in units of 10 N, then the maximum reading is X = 180500 N / 10 N/unit = 18050 units.
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