Question

For two waves, each of amplitude $A_0$ and frequency $f$, what is the resultant amplitude at a point where the phase difference between the waves is ${30}^{\circ }$?
Given, $\cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

• A. $A_0\sqrt{3}$
  
• B. $A_0\frac{\sqrt{3}}{2}$
  
• C. $A_0\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)$
  
• D. $A_0\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)$
  

Answer 1

The correct option is D $A_0\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)$


Given:
Amplitude of each wave $=A_0$
Phase difference between the waves,
$\Delta \varphi ={30}^{\circ }$

The resultant amplitude$\left(A\right)$ is given by,
$A=2A_0\cos \left(\frac{\Delta \varphi }{2}\right)$

$\Rightarrow A=2A_0\cos \left(\frac{{30}^{\circ }}{2}\right)$

Here, $\cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$\Rightarrow A=2A_0\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)=A_0\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)$

Hence, $\left(D\right)$ is the correct answer.

Remember! The value of $\cos {15}^{\circ }$ can be found by writing it as, $\cos {15}^{\circ }=\cos ({60}^{\circ }-{45}^{\circ })$ $=\cos {60}^{\circ }\cos {45}^{\circ }+\sin {60}^{\circ }\sin {45}^{\circ }$ and then substituting respective values in it.