JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2021
- Hard
- Numerical
Two waves are simultaneously passing through a string and their equations are : \({y}_{1}={A}_{1} \sin {k}({x}-v {t}), {y}_{2}={A}_{2} \sin {k}\left({x}-{vt}+{x}_{0}\right) .\) Given amplitudes \({A}_{1}=12\, {mm}\) and \({A}_{2}=5\, {mm}\) \({x}_{0}=3.5\, {cm}\) and wave number \({k}=6.28\, {cm}^{-1}\). The amplitude of resulting wave will be \(......\,{mm}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(7\)
Step-by-step Solution
Detailed explanation
\({y}_{1}={A}_{1} \operatorname{sink}({x}-{vt})\)
\({y}_{1}=12 \sin 6.28({x}-{vt})\)
\({y}_{2}=5 \sin 6.28({x}-{vt}+3.5)\)
\(\Delta \phi=\frac{2 \pi}{\lambda}(\Delta {x})\)
\(={K}(\Delta {x})\)
\(=6.28 \times 3.5=\frac{7}{2} \times 2 \pi=7 \pi\)
\({A}_{{DE}}=\sqrt{{A}_{1}^{2}+{A}_{2}^{2}+2 {A}_{1} {A}_{2} \cos \phi}\)
\({A}_{{DA}}=\sqrt{(12)^{2}+(5)^{2}+2(12)(5) \cos (7 \pi)}\)
\(=\sqrt{144+25-120}\)
\(=\,7\)
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