MCQ
The relation between time and displacement for two particles is given by
$y_1=0.06 \sin 2 \pi\left(0.04 t+\phi_1\right), y_2=0.03 \sin 2 \pi\left(1.04 t+\phi_2\right)$
The ratio of the intensity of the waves produced by the vibrations of the two particles will be
  • A
    $2: 1$
  • B
    $1: 2$
  • $4: 1$
  • D
    $1: 4$

Answer

Correct option: C.
$4: 1$
(c) $\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\left(\frac{0.06}{0.03}\right)^2=\frac{4}{1}$

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