Physics - Wave on String Question with Solution | TestHub

The given progressive waves are:

$Y_1=\sin 2\pi \left(\frac{t}{0.4}-\frac{x}{4}\right)$

$Y_2=\sin 2\pi \left(\frac{t}{0.4}+\frac{x}{4}\right)$

According to the principle of superposition, the resultant standing wave is given by:

$Y=Y_1+Y_2$

$Y=\sin 2\pi \left(\frac{t}{0.4}-\frac{x}{4}\right)+\sin 2\pi \left(\frac{t}{0.4}+\frac{x}{4}\right)$

Using the trigonometric identity $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$:

$Y=2\sin \left(\frac{2\pi t}{0.4}\right)\cos \left(-\frac{2\pi x}{4}\right)$

$Y=2\cos \left(\frac{\pi x}{2}\right)\sin \left(\frac{2\pi t}{0.4}\right)$

The amplitude of the standing wave at any position $x$ is given by the magnitude of the position-dependent part:

$A=\left|2\cos \left(\frac{\pi x}{2}\right)\right|$

Substituting $x=0.5\text{ m}$:

$A=\left|2\cos \left(\frac{\pi \times 0.5}{2}\right)\right|$

$A=2\cos \left(\frac{\pi }{4}\right)$

Since $\cos \left(\frac{\pi }{4}\right)=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$:

$A=2\times \frac{1}{\sqrt{2}}=\sqrt{2}\text{ m}$

Answer: $\sqrt{2}\mathrm{m}$