Question

Two progressive waves $Y_1=\sin 2\pi\left(\dfrac{t}{4}-\dfrac{x}{4}\right)$ and $Y_2=\sin 2\pi\left(\dfrac{t}{4}+\dfrac{x}{4}\right)$ superpose to form a standing wave. $Y_1$ and $Y_2$ are in SI system. Amplitude of the particle at $x=0.5\,\text{m$ is}
$\left[\sin 45^\circ=\cos 45^\circ=\dfrac{1}{\sqrt{2}}\right]$

• A. $2\sqrt{2}\,\text{m}$
• B. $2\,\text{m}$
• C. $\sqrt{2}\,\text{m}$
• D. $\dfrac{1}{\sqrt{2}}\,\text{m}$

Hint:

In standing waves, amplitude depends only on position, not on time.

Answer 1

The Correct Option is C

Step 1: Writing the resultant standing wave.
\[ Y = Y_1 + Y_2 = \sin 2\pi\left(\frac{t}{4}-\frac{x}{4}\right)+\sin 2\pi\left(\frac{t}{4}+\frac{x}{4}\right) \]
Step 2: Using trigonometric identity.
\[ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ Y = 2\sin\left(\frac{\pi t}{2}\right)\cos\left(\frac{\pi x}{2}\right) \]
Step 3: Expression for amplitude.
Amplitude at position $x$ is:
\[ A(x) = 2\left|\cos\left(\frac{\pi x}{2}\right)\right| \]
Step 4: Substituting $x=0.5\,\text{m$.}
\[ A = 2\cos\left(\frac{\pi}{4}\right)=2\cdot\frac{1}{\sqrt{2}}=\sqrt{2} \]
Step 5: Conclusion.
Amplitude of the particle is $\sqrt{2}\,\text{m}$.