Q103 / 160 · JIPMER 2005 · Physics
Equations of motion in the same direction are given by
$\begin{aligned}
& y_1=2 a \sin (\omega t-k x) \\
& y_2=2 a \sin (\omega t-k x-\theta)
\end{aligned}$
The amplitude of the medium particle will be
- A$2 a \cos \theta$
- B$\sqrt{2} a \cos \theta$
- Correct answerC$4 a \cos \frac{\theta}{2}$
- D$\sqrt{2} a \cos \frac{\theta}{2}$
Solution
The given equations of motion are
$\begin{aligned} & y_1=2 a \sin (\omega t-k x) \\ & y_2=2 a \sin (\omega t-k x-\theta)\end{aligned}$
Now the resultant equation of wave is given by
$\begin{aligned} & y=y_1+y_2 \\ & =2 a \sin (\omega t-k x)+2 a \sin (\omega t-k x-\theta) \\ & y=2 a\left[2 \sin \frac{(\omega t-k x+\omega t-k x-\theta)}{2}\times \cos \frac{\omega t-k x-(\omega t-k x-\theta)}{2}\right]\end{aligned}$
$y=4 a \cos \frac{\theta}{2} \sin \left(\omega t-k x-\frac{\theta}{2}\right)$ ...(1)
Now comparing equation (1) with $y=A \sin (\omega t-k x)$, we have
Resultant amplitude $A=4 a \cos \frac{\theta}{2}$
Hence, the correct answer is option: 'C'.