Question
A standing wave is represented by $\text{y}=2\text{A}\sin\text{kx}\cos\omega\text{t}.$ If one of the component waves is $\text{y}_1=\text{A}\sin(\omega\text{t}-\text{kx}),$ what is the equation of the second component wave?
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Answer
As $2\sin\text{A}\cos\text{B}=\sin(\text{A}+\text{B})+\sin(\text{A}-\text{B})$
$\text{y}=2\text{A}\sin\text{kx}\cos\omega\text{t}$
$\text{A}\sin(\text{kx}+\omega\text{t})+\text{A}\sin(\text{kx}-\omega\text{t})$
According to superposition principle,
$\text{y}=\text{y}_1+\text{y}_2;$
and $\text{y}_1=\text{A}\sin(\omega\text{t}-\text{kx})=-\text{A}\sin(\text{kx}-\omega\text{t})$
$\therefore \text{y}_2=\text{y}-\text{y}_1=2\text{A}\sin\text{kx}\cos\omega\text{t}+\text{A}\sin(\text{kx}-\omega\text{t})$
$=\text{A}\sin(\text{kx}+\omega\text{t})+2\text{A}\sin(\text{kx}-\omega\text{t})$
$=\text{a}\sin(\text{kx}+\omega\text{t})-2\text{A}\sin(\omega\text{t}-\text{kx}).$
$\text{y}=2\text{A}\sin\text{kx}\cos\omega\text{t}$
$\text{A}\sin(\text{kx}+\omega\text{t})+\text{A}\sin(\text{kx}-\omega\text{t})$
According to superposition principle,
$\text{y}=\text{y}_1+\text{y}_2;$
and $\text{y}_1=\text{A}\sin(\omega\text{t}-\text{kx})=-\text{A}\sin(\text{kx}-\omega\text{t})$
$\therefore \text{y}_2=\text{y}-\text{y}_1=2\text{A}\sin\text{kx}\cos\omega\text{t}+\text{A}\sin(\text{kx}-\omega\text{t})$
$=\text{A}\sin(\text{kx}+\omega\text{t})+2\text{A}\sin(\text{kx}-\omega\text{t})$
$=\text{a}\sin(\text{kx}+\omega\text{t})-2\text{A}\sin(\omega\text{t}-\text{kx}).$
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