Question 300: Waves

Question

Two progressive waves $Y_1=\sin 2 \pi\left(\frac{t}{0 \cdot 4}-\frac{x}{4}\right)$ and $Y_2=\sin 2 \pi\left(\frac{t}{0 \cdot 4}+\frac{x}{4}\right)$ superpose to form a standing wave. ' $x$ ' and ' $y$ ' are in SI system. Amplitude of the particle at $x=0.5 \mathrm{~m}$ is $\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$

Accepted Answer

To find the amplitude of the particle at $ x = 0.5 \, 	ext{m} $ resulting from the superposition of two progressive waves, we start by analyzing the given wave equations:
Given:
$    Y_1 = \sin \left(2 \pi \left( \frac{t}{0.4} - \frac{x}{4} ight) ight)    $
$    Y_2 = \sin \left(2 \pi \left( \frac{t}{0.4} + \frac{x}{4} ight) ight)    $
When these two waves superimpose, the resulting wave $ Y $ is given by:
$    Y = Y_1 + Y_2    $
Using the trigonometric identity for the sum of sines:
$    \sin(A + B) + \sin(A - B) = 2 \sin A \cos B    $
Applying this identity to simplify the expression:
$    Y = \sin \left(\frac{2 \pi t}{0.4} - \frac{2 \pi x}{4}ight) + \sin \left(\frac{2 \pi t}{0.4} + \frac{2 \pi x}{4}ight)    $
$    Y = 2 \sin \left(\frac{2 \pi t}{0.4}ight) \cos \left(\frac{2 \pi x}{4}ight)    $
Here, the amplitude $ R $ of the standing wave is:
$    R = 2 \cos \left(\frac{2 \pi x}{4}ight)    $
For $ x = 0.5 \, 	ext{m} $, calculate the amplitude $ R $:
$    R = 2 \cos \left(\frac{2 \pi 	imes 0.5}{4}ight) = 2 \cos \left(\frac{\pi}{4}ight) = 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2}    $
Thus, the amplitude of the particle at $ x = 0.5 \, 	ext{m} $ is $ \sqrt{2} \, 	ext{m} $.

Answer

$\sqrt{3} \mathrm{~m}$

Answer

$3 \sqrt{3} \mathrm{~m}$

Answer

$\sqrt{2} \mathrm{~m}$

Answer

$2 \sqrt{2} \mathrm{~m}$

Waves

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