A wave travels on a light string.The equation of the wave is Y = … - Vidyadip

MCQ

A wave travels on a light string.The equation of the wave is $Y = A \,\sin \,(kx - \omega t + 30^o)$.It is reflected from a heavy string tied to an end of the light string at $x = 0$. If $64\%$ of the incident energy is reflected the equation of the reflected wave is

A
$Y = 0.8\, A \sin \,(kx -\omega t + 30^o + 180^o)$
B
$Y = 0.8\, A \sin \,(kx + \omega t + 30^o + 180^o)$
✓
$Y = 0.8\, A \sin \,(kx + \omega t -30^o)$
D
$Y = 0.8\, A \sin \,(kx + \omega t + 30^o)$

✓

Answer

Correct option: C.

$Y = 0.8\, A \sin \,(kx + \omega t -30^o)$

c
$ y =.8 A \sin \left(-k x-\omega t+30^{\circ}+180^{\circ}\right) $

$=-.8 A \sin \left(-k x-\omega t+30^{\circ}\right) $

$=0.8 A \sin \left(k x+\omega t-30^{\circ}\right)$

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