JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 18)
Two waves are simultaneously passing through a string and their equations are :
y1 = A1 sin k(x $$-$$ vt), y2 = A2 sin k(x $$-$$ vt + x0). Given amplitudes A1 = 12 mm and A2 = 5 mm, x0 = 3.5 cm and wave number k = 6.28 cm$$-$$1. The amplitude of resulting wave will be ................ mm.
y1 = A1 sin k(x $$-$$ vt), y2 = A2 sin k(x $$-$$ vt + x0). Given amplitudes A1 = 12 mm and A2 = 5 mm, x0 = 3.5 cm and wave number k = 6.28 cm$$-$$1. The amplitude of resulting wave will be ................ mm.
Answer
7
Explanation
y1 = A1 sin k(x $$-$$ vt)
y1 = 12 sin 6.28 (x $$-$$ vt)
y2 = 5 sin 6.28 (x $$-$$ vt + 3.5)
$$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$$
$$ = K(\Delta x)$$
$$ = 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi $$
$${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$
$${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )} $$
$$ = \sqrt {144 + 25 - 120} $$
y1 = 12 sin 6.28 (x $$-$$ vt)
y2 = 5 sin 6.28 (x $$-$$ vt + 3.5)
$$\Delta \phi = {{2\pi } \over \lambda }(\Delta x)$$
$$ = K(\Delta x)$$
$$ = 6.28 \times 3.5 = {7 \over 2} \times 2\pi = 7\pi $$
$${A_{net}} = \sqrt {A_1^2 + A_2^2 + 2{A_1}{A_2}\cos \phi } $$
$${A_{net}} = \sqrt {{{(12)}^2} + {{(5)}^2} + 2(12)(5)\cos (7\pi )} $$
$$ = \sqrt {144 + 25 - 120} $$
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