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  • <img alt="y_1=a \sin k x \cos \omega t \text { and } y_2=a \sin (\omega t-k x) \text {}" src="https://entrancecorner.oncodecogs.com/gif.latex?y_1%3Da%20%5Csin%20k%20x%20%5Ccos%20%5Comega%20t%20%5Ct

y_1=a \sin k x \cos \omega t \text { and } y_2=a \sin (\omega t-k x) \text {}

The phase difference between two points x_1=\frac{\pi}{3 k} \text { and } x_2=\frac{3 \pi}{2 k}

are \phi_1 \text { and } \phi_2 for the two waves. The ratio \frac{\phi_1}{\phi_2} is

Option: 1

1


Option: 2

\frac{5}{6}


Option: 3

\frac{3}{4}


Option: 4

\frac{6}{7}


Answers (1)

best_answer

\text { At } x_1=\frac{\pi}{3 k} \text { and } x_2=\frac{3 \pi}{2 k} \text {, sin } k x_1 \text { or } k x_2 is not zero.

Therefore, neither of x_1 \text { or } x_2 is a node. 

\Delta x=x_2-x_1=\left(\frac{3}{2}-\frac{1}{3}\right) \frac{\pi}{k}=\frac{7 \pi}{6 k}

Since, \frac{2 \pi}{k}>\Delta x>\frac{\pi}{k}

Therefore, \phi_1=\pi \text { and } \phi_2=k \Delta x=\frac{7 \pi}{6}

\therefore \frac{\phi_1}{\phi_2}=\frac{6}{7}

Posted by

Deependra Verma

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