question_answer 48)

                  The displacement of an elastic wave is given by the function \[y=3\,\sin \,\omega t+4\,\cos \,\omega t\]                 Where \[y\] is in cm and t is in second. Calculate the resultant amplitude.

Answer:

                  \[y=3\,\sin \,\omega t+4\cos \,\omega t=3\,\sin \,\omega t+4\,\sin \]\[(\omega t-\pi /2)\]                 This wave is the result of the superposition of waves represented as \[{{y}_{1}}\,=3\,\sin \,\omega t\] and                 \[{{y}_{2}}\,=4\,\sin \,(\omega t-\pi /2)\]                 \[\therefore \] \[=\,a=\,3\,cm,\,\,b=4\,cm,\,\phi \,=\pi /2\]                 \[\therefore \] Resultant amplitude.                 \[A\,=\,\sqrt{{{a}^{2}}+\,{{b}^{2}}+\,2ab\,\,\cos \,\,\phi }\]                 \[\sqrt{9+16}\,=5\,cm.\]