Question Number 113243 by ShakaLaka last updated on 11/Sep/20
Commented by ShakaLaka last updated on 11/Sep/20
$$\mathrm{Can}\:\mathrm{anybody}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{question}? \\ $$
Commented by mr W last updated on 11/Sep/20
$$\mathrm{sin}\:\omega{t}\:\centerdot\:\mathrm{sin}\:\omega{at} \\ $$$${has}\:{no}\:{x}\:{as}\:{variable}! \\ $$$${please}\:{check}\:{the}\:{question}! \\ $$
Commented by ShakaLaka last updated on 11/Sep/20
$$\mathrm{I}\:\mathrm{will}\:\mathrm{have}\:\mathrm{to}\:\mathrm{ask}\:\mathrm{my}\:\mathrm{teacher} \\ $$$${that}\:{what}\:{is}\:{the}\:{correct}\:{question}. \\ $$$$\mathrm{because}\:\mathrm{this}\:\mathrm{question}\:\mathrm{I}\:\mathrm{wrote}\:\mathrm{is} \\ $$$$\mathrm{totally}\:\mathrm{same}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}. \\ $$$$\mathrm{Actually},\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{assignmemt},\:{which}\:{was} \\ $$$${written}\:{directly}\:{from}\:{university}, \\ $$$${not}\:{by}\:{my}\:{teacher}. \\ $$$${By}\:{the}\:{way},\:{there}\:{is}\:{also} \\ $$$${another}\:{mistake}\:{in}\:{the}\:{question} \\ $$$${that}\:{there}\:{is}\:{written}\:\frac{\partial^{\mathrm{2}} {v}}{\partial{c}^{\mathrm{2}} }\:{instead} \\ $$$${of}\:\frac{\partial^{\mathrm{2}} {v}}{\partial{t}^{\mathrm{2}} } \\ $$
Answered by mathdave last updated on 12/Sep/20
$${if}\:{only}\:{u}\left({x},{t}\right)=\mathrm{sin}{wx}.\mathrm{sin}{at} \\ $$$${then}\:{is}\:{a}\:{solution}\:{that}\:{satisfy}\:{the}\:{wave} \\ $$$${equation}\:\frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} }={a}^{\mathrm{2}} .\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} } \\ $$$${by}\:{letting} \\ $$$$\frac{\partial^{\mathrm{2}} \left(\mathrm{sin}{wx}.\mathrm{sin}{wat}\right)}{\partial{t}^{\mathrm{2}} }=−{w}^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sin}{wx}.\mathrm{sin}{wat} \\ $$$${and} \\ $$$${a}^{\mathrm{2}} .\frac{\partial^{\mathrm{2}} \left(\mathrm{sin}{wx}.\mathrm{sin}{wat}\right)}{\partial{x}^{\mathrm{2}} }=−{w}^{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{sin}{wx}.\mathrm{sin}{wat} \\ $$$$\because{u}\left({x},{t}\right)=\mathrm{sin}{wx}.\mathrm{sin}{wat}\:{is}\:{a}\:{solution}\:{to} \\ $$$${wave}\:{equation}\: \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{t}^{\mathrm{2}} }={a}^{\mathrm{2}} .\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$
Commented by ShakaLaka last updated on 13/Sep/20
$${yeah}\:{mathdave},\:{it}\:{can}\:{be} \\ $$$${possible} \\ $$
Commented by ShakaLaka last updated on 15/Sep/20
$${mathdave}\:{you}\:{were}\:{right}.\:{my} \\ $$$${teacher}\:{is}\:{also}\:{saying}\:{that}\:{the} \\ $$$${function}\:“\mathrm{sin}\:\omega{t}\:\centerdot\:\mathrm{sin}\:\omega{a}\mathrm{t}''\:{is} \\ $$$${incorrect},\:{the}\:{correct}\:{function} \\ $$$${is}\:“\mathrm{sin}\:\omega{x}\:\centerdot\:\mathrm{sin}\:\omega{a}\mathrm{t}'';\:{as}\:{you}\:{said}. \\ $$