Question

Two waves are travelling in the same direction along a stretched string. The waves are ${90}^0$ out of phase. Each wave has an amplitude of 4.0 cm. The amplitude of the resultant wave is $5.66\times {10}^{-x}m$. Find x.

• A. 1
• B. 2
• C. 8
• D. 4

Answer 1

The correct option is B 2

Let A be the amplitude of each wave. Thus the waves are

$y_1=Acos(\omega t-kx)$

and $y_2=Asin(\omega t-kx)$

Thus the net wave is $y=y_1+y_2=Acos(\omega t-kx)+Asin(\omega t-kx)$

$=\sqrt{2}Asin(\omega t-kx+\frac{\pi }{4})$

Thus the new amplitude is $\sqrt{2}A=5.66cm=5.66\times {10}^{-2}m$

Thus, x=2.