JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 23)
The displacement equations of two interfering waves are given by
$y_{1}=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}, y_{2}=5[\sin \omega t+\sqrt{3} \cos \omega t] \mathrm{cm}$ respectively.
The amplitude of the resultant wave is _______ $\mathrm{cm}$.
$y_{1}=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}, y_{2}=5[\sin \omega t+\sqrt{3} \cos \omega t] \mathrm{cm}$ respectively.
The amplitude of the resultant wave is _______ $\mathrm{cm}$.
Answer
20
Explanation
Given, $y_{1}=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{cm}$ and
$y_{2}=5(\sin \omega t+\sqrt{3} \cos \omega t)$
$$ =10 \sin \left(\omega t+\frac{\pi}{3}\right) $$
Thus the phase difference between the waves is 0 .
$$ \text { so } A=A_{1}+A_{2}=20 \mathrm{~cm} $$
$y_{2}=5(\sin \omega t+\sqrt{3} \cos \omega t)$
$$ =10 \sin \left(\omega t+\frac{\pi}{3}\right) $$
Thus the phase difference between the waves is 0 .
$$ \text { so } A=A_{1}+A_{2}=20 \mathrm{~cm} $$
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