Waves Test - 63

$$\begin{array}{l} m{ w^{ 2 } }=k \\ Total\, \, initial\, \, energy=\frac { 1 }{ 2 } k{ A^{ 2 } } \\ at\, x=\frac { { 2A } }{ 3 } ,\, \, potential\, \, energy=\frac { 1 }{ 2 } k{ \left( { \frac { { 2A } }{ 3 }  } \right) ^{ 2 } } \\ =\left( { \frac { 1 }{ 2 } k{ A^{ 2 } } } \right) \left( { \frac { 4 }{ 9 }  } \right)  \\ Kinetic\, \, energy\, \, at\left( { x=\frac { { 2A } }{ 3 }  } \right) \, \, is=\left( { \frac { 1 }{ 2 } k{ A^{ 2 } } } \right) \left( { \frac { 5 }{ 9 }  } \right)  \\ If\, \, speed\, \, is\, \, triled,\, \, new\; \; kinetic\, \, energy=\frac { 1 }{ 2 } k{ A^{ 2 } }\cdot \frac { 5 }{ 9 } =\frac { 5 }{ 2 } k{ A^{ 2 } } \\ \therefore New\, \, total\, \, energy\, \, =\frac { 5 }{ 2 } k{ A^{ 2 } }+\frac { 1 }{ 2 } k{ A^{ 2 } }\left( { \frac { 4 }{ 9 }  } \right)  \\ if\, \, next\, \, amplitude\, \, =A' \\ then\, \, \frac { 1 }{ 2 } kA{ '^{ 2 } }\left( { \frac { { 49 } }{ 9 }  } \right)  \\ \Rightarrow A'=\frac { { 7A } }{ 3 }  \end{array}$$

For resonance, amplitude must be maximum which is possible only when the denominator of expression is zero i.e.
$$a \omega^2-b \omega +c=0 $$

 

The correct option is D

Hint: Using trigonometry, convert the given function in suitable form.

Explanation for correct answer:

Step 1: Information required,

The displacement equation is given as,

$$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\cos \omega t$$

Reworking the wave condition as $$\sin $$. Change the $$\cos $$ to $$\sin $$ by $$\cos \omega t$$ = $$ $$ $$\sin \left( {\omega t + \dfrac{\pi}{2}} \right)$$

$$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)………….(1)$$

From the above equation it is observed that the phase difference is $$\dfrac{\pi }{2}$$.

Step 2: Consider general wave equation,

The general wave equation is given as,

$$y = A\sin \left( {\omega t + \theta } \right)…………….(2)$$

Where A represents the amplitude.

Compare amplitudes in both the equations (1) and (2),

Consider amplitudes as $${A_{1,}}{A_2}$$

Then,

$${A_1} = \dfrac{1}{{\sqrt a }}$$  ,  $${A_2} = \dfrac{1}{{\sqrt b }}$$

Resultant amplitude is given as,

 $$A = \sqrt {{A_1}^2 + {A_2}^2} $$

Substituting the $${A_{1,}}{A_2}$$  values in the above equation, we get,

$$A = \sqrt {{{(\frac{1}{{\sqrt a }})}^2} + {{(\frac{1}{{\sqrt b }})}^2}} $$

 $$ \Rightarrow A = \sqrt {\dfrac{1}{a} + \dfrac{1}{b}}$$  

$$\therefore A = \sqrt {\dfrac{{a + b}}{{ab}}} \\ $$

Therefore, the amplitude of the wave is $$ A = \sqrt {\dfrac{{a + b}}{{ab}}} \\ $$.

$$\begin{array}{l} If\, \, the\, particle\, \, performs\, \, SHM\, \, with\, \, amplitude\, \, A\, \, along\, \, the\, \, straight\, \, line,\, \, than\, \, at\, \, any\, \, po{ { int } }\, \, the\, \, total\, \, energy\, \, should\, \, be, \\ E=K+P=\frac { 1 }{ 2 } K{ A^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } } \\ If\, \, the\, additional\, \, kinetic\, \, energy\, \, of\, \, magnitude\, \, \frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }\, \, by\, \, means\, \, of\, \, an\, \, impulsive\, \, force, \\ then\, \, the\, \, total\, \, energy\, \, of\, \, the\, \, particle\, \, changes\, \, to, \\ E=E+\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } }+\frac { 1 }{ 2 } m{ w^{ 2 } }{ A^{ 2 } } \\ \frac { 1 }{ 2 } m{ w^{ 2 } }A{ '^{ 2 } }=\frac { 1 }{ 2 } m{ w^{ 2 } }\left( { 2{ A^{ 2 } } } \right)  \\ \Rightarrow A{ '^{ 2 } }=2{ A^{ 2 } } \\ \Rightarrow A'=\sqrt { 2 } A \end{array}$$

Hence, option $$(C)$$ is correct answer.

Given,

$$\begin{array}{l} Given, \\ Dis\tan  ce\, \, x=\frac { { \sqrt { 3 }  } }{ 2 }  \\ Increased\, kinetic\, energy\, \, K.E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ now, \\ E=K.E+P.E \\ E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ x^{ 2 } }+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }\left( { { a^{ 2 } }-{ x^{ 2 } } } \right)  \\ E=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ When\, particle\, is\, at\, a\, dis\tan  ce\, of\, \frac { { \sqrt { 3 }  } }{ 2 } a\, from\, mean\, position,\, its\, kinetic\, energy\, gets\,  \\ increased\, by\, an\, amount\, \frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } }\, due\, to\, an\, impulsive\, force. \\ So,\, the\, new\, total\, energy\, is\,  \\ E'=E+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ \frac { 1 }{ 2 } m{ \omega ^{ 2 } }a{ '^{ 2 } }=\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } }+\frac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } \\ a'=\sqrt { 2 } a \\ Hence,\; the\, option\, C\, is\, the\, correct\, answer. \end{array}$$

$$\begin{array}{l} We\, have \\ x=A\sin  \left( { \omega t+\delta  } \right) ;\, \, where\, \delta \, is\, the\, phase\, cons\tan  t. \\ For\, particle\, 1 \\ at\, X=A\, and\, t=0 \\ \frac { A }{ 2 } =A\sin  \delta  \\ \Rightarrow \frac { 1 }{ 2 } =\sin  \delta  \\ \Rightarrow \delta =\frac { \pi  }{ 6 } ,\frac { { 5\pi  } }{ 6 }  \\ The\, velocity\, is \\ v=\frac { { dx } }{ { dt } } =A\, \omega \cos  \left( { \omega t+\delta  } \right)  \\ At\, t=0\, ,\, v=A\omega t\cos  \delta ]Now, \\ \cos  \left( { \frac { \pi  }{ 6 }  } \right) =\frac { { \sqrt { 3 }  } }{ 2 } \, and\, \cos  \left( { \frac { { 5\pi  } }{ 6 }  } \right) =-\frac { { \sqrt { 3 }  } }{ 2 }  \\ As\, v\, is\, positive\, for\, particle\, 1,\delta \, must\, be\, equal\, to\, \frac { \pi  }{ 6 }  \\ for,\, particle\, 2,\, V\, is\, negative\, ,\, \delta \, must\, be\, equal\, to\, \frac { { 5\pi  } }{ 6 }  \\ Thus,\, the\, phase\, difference\, between\, them\, is\, =\frac { { 5\pi  } }{ 6 } -\frac { \pi  }{ 6 }  \\ =\frac { { 4\pi  } }{ 6 }  \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$$