The relation between time and displacement for two particles is g… - Vidyadip

MCQ

The relation between time and displacement for two particles is given by

${y_1} = 0.06\sin 2\pi (1.04t + {\phi _1})$,

${y_2} = 0.03\sin 2\pi (1.04t + {\phi _2})$

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
(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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