A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant \(0.5\ \mathrm{N\,m^{-1}}\) (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillation it performs before its amplitude decreases by a factor of \(e\) is \(N\). If the mass of the strip is 50 grams, its resistance \(10\ \Omega\) and air drag negligible, \(N\) will be close to :
Show Hint
- 50000
- 1000
- 5000
- 10000
The Correct Option is C
Solution and Explanation
\(b = \frac{(0.1)^2(0.1)^2}{10} = 10^{-5}\ \mathrm{kg/s}\).
For damped SHM the amplitude is \(A = A_0e^{-bt/2m}\); it falls by a factor \(e\) at \(t = \frac{2m}{b} = \frac{2\times 0.05}{10^{-5}} = 10^4\ \mathrm{s}\).
Time period \(T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.05}{0.5}} = 2\pi\sqrt{0.1} \approx 1.99\ \mathrm{s}\) (damping is very weak, so \(T\) is unchanged).
\(N = \frac{10^4}{1.99} \approx 5030 \approx 5000\).
Answer: (3) 5000
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