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This is one of our many quizzes on Oscillations & waves. Click on this to access more such competitive quizzes on Waves.

Oscillations & waves Quiz 2

physics

 The three-dimensional wave equation is given by ${\nabla ^2}\phi  = \dfrac{1}{{{\upsilon ^2}}}\dfrac{{{\partial ^2}\phi }}{{\partial {t^2}}}$. Here $\phi $ represents the vector representation of a harmonic plane wave, is given by the equation?

 The vector representation of a harmonic plane wave in three dimensions is represented by $\phi $, where in $\vec{k}$ is the vector along the direction of wave propagation, given by the equation $\left| {\vec k} \right| = \dfrac{{2\pi }}{\lambda }$. 

  Given; $\\ \Rightarrow {{\nabla }^{2}}\phi =\dfrac{1}{{{\upsilon }^{2}}}\dfrac{{{\partial }^{2}}\phi }{\partial {{t}^{2}}} \\ $ Where ${{\nabla }^{2}} $ is the Laplacian operator. $\\ $ Here, $\phi$ is the vector representation of a harmonic plane wave in three dimensions. It is given by: $ \\ \phi =A\sin \left( \vec{k}.\vec{r}-\omega t \right) \\ $ where, $\vec{k}$ is the vector along the direction of propagation of the wave, known as propagation vector, $\left| {\vec{k}} \right|=\dfrac{2\pi }{\lambda}$

 A squeezing wave of frequency $420{\rm{ }}Hz$ is trapped in an iron rod and then passed from the iron rod into air. The speed of the wave is $6400{\rm{ }}m/s$ in iron and $480{\rm{ }}m/s$ in air. Find the length of the wave both in iron rod and in air. 

 As the frequency of the wave remains constant while passing from one medium to another, so, you can use the wavelength formula, $\lambda  = \dfrac{\upsilon }{v}$, to get the answer.

  The frequency of a wave remains unchanged as it passes from one medium to another. In iron, $ \\ \lambda =\dfrac{\upsilon }{v}=\dfrac{6400}{420}=15.2\,m \\ \text{In air}, \\ \lambda =\dfrac{\upsilon }{v}=\dfrac{480}{420}=1.1\,m$

 Considering the Poisson bracket as a dynamical quantity, which of the following statements are true regarding the poisson bracket? 

 While considering the two dynamical quantities of the poisson brackets you will find that the poisson bracket and the Hamiltonian function appear in a certain way. Also, the poisson bracket has a unique system of function of derivatives that correlates to its coordinates and the momenta. 

 The poisson bracket is a skew-symmetric bilinear of derivatives. It has two dynamical quantities that appear with respect to the coordinates and the momenta. A dynamical quantity can be any physically relevant function, such as: $ \\ $ Kinetic energy, Hamiltonian function or angular momentum $ \\ $ The poisson bracket of $g$ and the Hamiltonian function appear in a natural way if we calculate the total time change of $g$, along a physical orbit in phase space.

 Assume that two symmetrical objects are in symmetry along the axis of rotation. Then which among the following statements would be true, if we are to consider the total momentum of the object?

 Consider the total angular momentum in terms of its angular velocity and the moment of inertia. This will correspond to the particle's velocity and the perpendicular distance along the axis of rotation.

 The component of angular momentum parallel to fixed axis, along z-axis is given by: $ \\ L=\sum{I=\sum{({{I}_{p}}+{{I}_{z}})}} \\ $ Also, every particle possess a velocity $vi$ having a corresponding particle possessing velocity $-vi$. $ \\ $ The object under consideration is having a general symmetry. Due to this symmetry about axis of rotation, it also possess a particular perpendicular distance $rp$ due to which, the total angular momentum of these particles cancel each other. The relation: $ \\ L={{L}_{z}}={{I}_{z}}\omega {k}' \\ $  This gives the total angular momentum of these particles. Where, $\omega$ is the angular velocity and, $I$ is the moment of inertia of the object.

 The mass on a vertical vault undergoes forced vibrations according to the equation $\ddot x + {\omega ^2}x = f\sin \omega t$, where there is no damping and the imposed frequency is equal to the natural frequency of oscillation. Find the solution of the above differential equation. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a6aa4cdf-5152-43ef-8ca9-79e8bcb2c6cd667182896365820576.png"/>

 The general solution can be obtained by making an equation of the form $x = {x_1} + {x_2}$ , using which we can obtain the particular integral of the equation, which can be substituted and simplified to obtain the solution.

 $\text{The general solution of the equation:} \\ \ddot{x}+{{\omega }^{2}}x=f\sin \omega t\text{            }...\text{(i)} \\ \text{is, }x={{x}_{1}}+{{x}_{2}}\text{                    }...\text{(ii)} \\ {{x}_{1}}\text{ is the general solution of the homogeneous equation:} \\ \Rightarrow {{{\ddot{x}}}_{1}}+{{\omega }^{2}}{{x}_{1}}=0\text{                }...\text{(iii)} \\ {{x}_{2}}\text{ is the particular integral equation of general solution}\text{.} \\ \text{The general solution for the equation is:} \\ {{x}_{1}}=A\cos \omega t+B\sin \omega t\text{     }...\text{(iv)} \\ \text{The particular solution of the equation (i) is given by:} \\ x=t\left[ {{c}_{1}}\cos \omega t+{{c}_{2}}\sin \omega t \right]\text{ } \\ \text{which gives:} \\ \dot{x}=\left( {{c}_{1}}\cos \omega t+{{c}_{2}}\sin \omega t \right)+t\left( -\omega {{c}_{1}}\sin \omega t+\omega {{c}_{2}}\cos \omega t \right) \\ \ddot{x}=2\left( -\omega {{c}_{1}}\sin \omega t+\omega {{c}_{2}}\cos \omega t \right)-t{{\omega }^{2}}\left( {{c}_{1}}\cos \omega t+{{c}_{2}}\sin \omega t \right) \\ \text{Substituting these equations into (i), we get,} \\ 2{{c}_{2}}\omega \cos \omega t-2{{c}_{1}}\omega \cos \omega t=f\sin \omega t \\ \text{from which }{{c}_{2}}=0\text{ and }{{c}_{1}}=-\dfrac{f}{2\omega }. \\ \text{Thus, the particular integral is:} \\ {{x}_{2}}=-\dfrac{ft}{2\omega }\cos \omega t \\ \text{The general solution of equation (i) is,} \\ x=A\cos \omega t+B\sin \omega t-\dfrac{ft}{2\omega }\cos \omega t$