Question
A wave $\text{Y} = \text{A}\sin(\omega\text{t} - \text{kx})$ allowed through a string gets reflected from a rigid support and forms stationary wave. Derive the expression for the standing wave.
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Answer
Given wave: $\text{Y}_\text{i}=\text{A}\sin(\omega\text{t}-\text{Kx})$
Reflected wave:
$\text{Y}_\text{r}=\text{A}\sin (\omega\text{t}+\text{Kx}+\pi)$
$=-\text{A}\sin(\omega\text{t}+\text{Kx})$
Applying superposition principle,
$\text{y}=\text{Y}_\text{i}+\text{Y}_\text{r}$
$=\text{A}\sin(\omega\text{t}-\text{Kx})-\text{A}\sin(\omega\text{t}+\text{Kx})$
$\text{y}=2\text{A}\sin\text{Kx}\cos\omega\text{t}$
Since amplitude $2\text{A}\sin\text{Kx}$ varies with position, it represents a standing wave.
Reflected wave:
$\text{Y}_\text{r}=\text{A}\sin (\omega\text{t}+\text{Kx}+\pi)$
$=-\text{A}\sin(\omega\text{t}+\text{Kx})$
Applying superposition principle,
$\text{y}=\text{Y}_\text{i}+\text{Y}_\text{r}$
$=\text{A}\sin(\omega\text{t}-\text{Kx})-\text{A}\sin(\omega\text{t}+\text{Kx})$
$\text{y}=2\text{A}\sin\text{Kx}\cos\omega\text{t}$
Since amplitude $2\text{A}\sin\text{Kx}$ varies with position, it represents a standing wave.
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