Two progressive waves $Y_1=\sin 2 \pi\left(\frac{t}{0 \cdot 4}-\frac{x}{4}\right)$ and $\mathrm{Y}_2=\sin 2…

Two progressive waves $Y_1=\sin 2 \pi\left(\frac{t}{0 \cdot 4}-\frac{x}{4}\right)$ and $\mathrm{Y}_2=\sin 2 \pi\left(\frac{\mathrm{t}}{0 \cdot 4}+\frac{\mathrm{x}}{4}\right)$ superpose to form a standing wave. ' $x$ ' and ' $y$ ' are in SI system. Amplitude of the particle at $x=0.5 \mathrm{~m}$ is $\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$
  1. $\sqrt{3} \mathrm{~m}$
  2. $3 \sqrt{3} \mathrm{~m}$
  3. $\sqrt{2} \mathrm{~m}$
  4. $2 \sqrt{2} \mathrm{~m}$

Solution

$Y=Y_1+Y_2$ $\begin{array}{ll}\therefore \quad & Y=\sin \left(\frac{2 \pi t}{0.4}-\frac{2 \pi x}{4}\right)+\sin \left(\frac{2 \pi t}{0.4}+\frac{2 \pi x}{4}\right) \\ & \sin (A+B)+\sin (A-B)=2 \sin A \cos B \\ \therefore \quad & Y=2 \sin \left(\frac{2 \pi t}{0.4}\right), \cos \left(\frac{2 \pi x}{4}\right) \\ & Y=R \sin \left(\frac{2 \pi t}{0.4}\right), \text { where } R \text { is the amplitude } \\ \therefore \quad & R=2 \cos \left(\frac{2 \pi x}{4}\right) \\ \therefore \quad & \text { at } x=0.5 \\ & R=2 \cos \left(\frac{2 \pi}{4 \times 2}\right)=2 \cos \left(\frac{\pi}{4}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}\end{array}$

Asked in: MHT CET 2024 (15 May Shift 1)

Practice more Waves and Sound questions on Aicharya