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  2. Can someone explain - Oscillations and Waves - JEE Main-7
# Engineering

A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connect to a spring of spring constant 0.5\; Nm^{-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 oscillations 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 :

  • Option 1)

    10000

  • Option 2)

    1000

  • Option 3)

    5000

  • Option 4)

    50000

Answers (2)
S solutionqc

l=10 \; cm

k=0.5\; N/m

B=0.1\; T

-kx-ilB = \frac{md^{2}x}{dt^{2}}.....\; \; \; \; (1)

i =\frac{Bl}{R}\frac{dx}{dt} \; \; \; \; \; \; (2)\rightarrow \; Put\; in \left ( 1 \right )

-kx-\frac{B^{2}l^{2}}{R}\frac{dx}{dt}=\frac{md^{2}x}{dt^{2}}

\Rightarrow A=A_{o}e^{-kt}

for A=\frac{A_{o}}{e}=A_{o}e^{-kt}

kt=1

\Rightarrow \left ( \frac{B^{2}l^{2}}{R.2m} \right )t=1

t=\frac{2mR}{B^{2}l^{2}}=\frac{2\times 50\times 10^{-3}\times 10}{0.1\times 0.1\times 10\times 10^{-4}\times 10}

t=10^{4}s

T_{o}=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{50\times 10^{-3}}{0.5}}=\frac{2\pi }{\sqrt{10}}

No. of oscillations =\frac{t}{T_{o}}=\frac{10^{4}\times \sqrt{10}}{2\pi }\approx 5000


Option 1)

10000

Option 2)

1000

Option 3)

5000

Option 4)

50000

A Ashutosh

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