Waves Test - 42

Given, $$y_{1} = a_{1}\sin \left (\omega t - \dfrac {2\pi x}{\lambda}\right )$$ and
$$y_{2} = a_{2} \cos \left (\omega t - \dfrac {2\pi x}{\lambda} + \phi \right )$$
$$= a_{2}\sin \left (\dfrac {\pi}{2} + \omega t - \dfrac {2\pi x}{\lambda} + \phi \right )$$
$$\therefore$$ Phase of first wave, $$\phi_{1} = \left (\omega t - \dfrac {2\pi x}{\lambda}\right )$$
and phase of second wave, $$\phi_{2} = \dfrac {\pi}{2} + \omega t - \dfrac {2\pi x}{\lambda} + \phi$$
$$\therefore$$ Phase difference, $$\triangle \phi = \phi_{2} - \phi_{1} = \dfrac {\pi}{2} + \phi$$
$$\therefore$$ Path difference $$= \dfrac {\lambda}{2\pi}\times phase\ difference$$
$$= \dfrac {\lambda}{2\pi}\left (\dfrac {\pi}{2} + \phi \right )$$.

$$y_1 = \dfrac{1}{2} sim \omega t + \dfrac{\sqrt 3}{2} cos \omega t$$
Thus we get  $$y_1= cos \dfrac{\pi}{3} sim \omega t + sin \dfrac{\pi}{3} cos \omega t$$
$$\therefore y_1 = sim (\omega t + \pi/3)$$
$$y_2 = \sqrt 2 \left( \dfrac{1}{\sqrt 2} sim \omega t + \dfrac{1}{\sqrt 2} cos  \omega t\right )$$
Similarly, $$y_2 = \sqrt 2 sin (\omega t + \pi/4)$$
Phase difference $$= \Delta \Phi = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$$

Let phase difference between them is $$\phi$$ .
So, $$x_1=a\,sin \,\omega t $$  and  $$ x_2=9\,sin (\omega t+\phi)$$
or     $$\dfrac{a}{3}=a\,sin\,\omega t$$
$$\Rightarrow$$   $$sin \,\omega t =\dfrac{1}{3}$$...............(i)
Similarly , $$-\dfrac{a}{3}= a\,sin\,(\omega t+\phi)$$
$$sin\,(\omega t +\phi)=-\dfrac{1}{3}$$...................................(ii)
Eliminating $$t,$$ we get 
$$ cos \,\phi=-1\,,\dfrac{7}{9}$$
$$\Rightarrow$$ $$\phi=180^{0}$$
or $$\phi=cos^{-1}\,(\dfrac{7}{9})$$

The amplitude for first SHM equation, $$A_1=\sqrt{5^2}=5$$ and 

Given, $$y_{1} = A\sin \omega t$$
and $$y_{2} = \dfrac {A}{2} \sin \omega t + \dfrac {A}{2} \cos \omega t$$
or $$y_{2} = \dfrac {A}{2} (\sin \omega t + \cos \omega t)$$
$$= \dfrac {A}{3} \sqrt {2} [\sin (\omega t + 45^{\circ}]$$
$$= \dfrac {A}{\sqrt {2}} \sin (\omega t + 45^{\circ})$$
$$\therefore$$ The ration of amplitudes of two motions
$$\dfrac {A_{1}}{A_{2}} = \dfrac {A}{A\sqrt {2}} = \sqrt {2}$$.

Given equation can be written as $$y = 6\sin(4\pi t- 0.2\pi x)$$

At the same depth under water velocity of sound will be same 
$$\therefore$$    $$v_1=v_2$$
$$\Rightarrow$$ $$\dfrac{\lambda_1}{T_1}=\dfrac{\lambda_2}{T_2}$$
$$\therefore$$   $$\lambda_2=\dfrac{T_2}{T_1}\lambda_1$$
$$=\dfrac{1/2}{1/10} \times 30 =150 mm$$

Let equation of motion to particles be given by,

Lets first find resultant of four waves using phase diagram

first wave is representated along$$+ve$$ X-axis with vector of length equal to amplitude of the wave.

resultant of those four can be written as (by adjusting X and Y component)

$$R=\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =4\sqrt { 2 } $$

and $$\sin { \theta =\cfrac { 4 }{ 4\sqrt { 2 }  }  } =\cfrac { 1 }{ \sqrt { 2 }  } \\ \therefore \theta =\cfrac { \pi  }{ 4 } \\ \therefore { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }=4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi  }{ 4 }  \right)  } $$

now add fifth wave

$$\therefore y=\left( { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } \right) +{ y }_{ 5 }\\ =4\sqrt { 2 } \sin { \left( wt+kx+\cfrac { \pi  }{ 4 }  \right)  } +4\sqrt { 2 } \sin { \left( wt-kx+\cfrac { \pi  }{ 4 }  \right)  } \\ =4\sqrt { 2 } \left[ 2\times \sin { \left( wt+\cfrac { \pi  }{ 4 }  \right) cos\left( kx \right)  }  \right] $$

By using $$\sin { C } +\sin { D } =2\sin { \cfrac { C+D }{ 2 } \cos { \cfrac { C+D }{ 2 }  }  } $$

$$\therefore y=8\sqrt { 2 } \sin { \left( wt+\cfrac { \pi  }{ 4 }  \right)  } \cos { \left( kx \right)  } $$