Question 324: Waves

Question

The equations of two waves are given as
$$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1ight) \
& y_2=a \sin \left(\omega t+\phi_2ight)
\end{aligned}$$
If amplitude and time period of resultant wave is same as the individual waves, then $\left(\phi_1-\phi_2ight)$ is

Accepted Answer

Let's start by writing down the two wave equations:
$$y_1 = a \sin (\omega t + \phi_1)$$
$$y_2 = a \sin (\omega t + \phi_2)$$
When we add these two waves, we can use the trigonometric identity:
$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}ight) \cos\left(\frac{A-B}{2}ight)$$
Letting
$$A = \omega t + \phi_1 \quad 	ext{and} \quad B = \omega t + \phi_2,$$
the sum becomes:
$$y = y_1 + y_2 = 2a \cos\left(\frac{\phi_1 - \phi_2}{2}ight) \sin \left(\omega t + \frac{\phi_1 + \phi_2}{2}ight)$$
This represents a single sinusoidal wave with:
Amplitude: $$A' = 2a \left|\cos\left(\frac{\phi_1 - \phi_2}{2}ight)ight|$$
Angular frequency: $$\omega$$ (and hence the same time period)
The problem states that the amplitude of the resultant wave is the same as that of the individual waves. Thus, we require:
$$2a \left|\cos\left(\frac{\phi_1 - \phi_2}{2}ight)ight| = a$$
Dividing both sides by $$a$$ (assuming $$a 
eq 0$$), we get:
$$2\left|\cos\left(\frac{\phi_1 - \phi_2}{2}ight)ight| = 1$$
This simplifies to:
$$\left|\cos\left(\frac{\phi_1 - \phi_2}{2}ight)ight| = \frac{1}{2}$$
The cosine function equals $$\frac{1}{2}$$ when its argument is $$\frac{\pi}{3}$$ (or $$-\frac{\pi}{3}$$, but the absolute value takes care of the sign). Similarly, $$\cos\left(\frac{\phi_1 - \phi_2}{2}ight)$$ equals $$-\frac{1}{2}$$ when the argument is $$\frac{2\pi}{3}$$ (or $$-\frac{2\pi}{3}$$). In either case, the magnitude is $$\frac{1}{2}$$.
Choosing the positive difference for simplicity, we set:
$$\frac{\phi_1 - \phi_2}{2} = \frac{\pi}{3} \quad \Rightarrow \quad |\phi_1 - \phi_2| = \frac{2\pi}{3}$$
This result can be written as:
$$\phi_1 - \phi_2 = \cos^{-1}\left(-\frac{1}{2}ight)$$
since
$$\cos\left(\frac{2\pi}{3}ight) = -\frac{1}{2}.$$
Thus, the correct answer is Option A:
$$\cos^{-1}\left(\frac{-1}{2}ight).$$

Answer

$\cos ^{-1}\left(\frac{-1}{2}\right)$

Answer

$\cos ^{-1}\left(\frac{-1}{4}\right)$

Answer

$\cos ^{-1}\left(-\frac{1}{6}\right)$

Answer

$\cos ^{-1}\left(-\frac{1}{8}\right)$

Waves

The equations of two waves are given as $\begin{aligned} & y_1=a \sin \left(\omega t+\phi_1\right) \\ & y_2=a \sin \left(\omega t+\phi_2\right) \end{aligned} If amplitude and time period of resultant wave is same as the individual waves, then \left(\phi_1-\phi_2\right)$ is

Choose the correct answer: