Oscillations - Class 11 Medical Physics - Extra Questions

$$R= \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos( \alpha_1 - \alpha_2)}$$

$$\phi = \tan^{-1} \left[ \dfrac{a_1 \sin \alpha_1 + a_2 \sin \alpha_2}{a_1 \cos \alpha_1 + a_2 \cos \alpha_2} \right]$$

$$x=kt\\ v=\cfrac { dx }{ dt } =k\\ a=\cfrac { dv }{ dt } =0$$

$$\therefore$$ Equation of SHM can be written as $$\displaystyle x-4=2\cos \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{2}{2}}=1$$
i.e. x = 4 + 2 cos t

Given, $$x=5m$$, $$t=0.1s$$
$$Y=0.05\sin { \pi  } \left[ 20t-\cfrac { x }{ 6 }  \right] $$
$$\therefore$$ Displacement $$Y=0.05\sin { \pi  } \left[ 20\times 0.1-\cfrac { 5 }{ 6 }  \right] $$
$$=0.05\sin { \pi  } \left( 2-0.833 \right) =0.025m$$

The general equations of a simple harmonic motion,

Given, $$x= (2.0 cm) sin[ 100 t + \dfrac{\pi}{6}]$$

For equilibrium , $$m\omega^2r=mg\cos\theta$$

The equation of motion of S.H.M $$x=0.5\cos { (100+\cfrac { 3\pi  }{ 4 } ) } $$. Comparing it with general equation motion of S.H.M, described as follows, we get, $$x=A\cos { (\omega t+\theta ) } $$

No, every oscillatory motion is simple harmonic motion.

Given,

Position function, $$X=A\sin (\pi t+\phi )\,......\,\,(1)$$

Velocity function, $$v=\dfrac{dX}{dt}=A\pi \cos (\pi t+\phi )\,.....\,...\,(2)$$

Condition, (at $$t=0,$$ velocity $$v=\pi \,cm{{s}^{-1}}$$ and$$x=1\,cm$$ )

From equation (1),   $$1=A\sin (\pi \times 0+\phi )\,\,......\,\,(3)$$

From equation (2),  $$\pi =A\pi \cos (\pi \times 0+\phi )\,\,........\,\,(4)$$

From (3) and (4)

$$\tan \phi =1\,\,\,\Rightarrow \,\,\,\,\phi =\pi /4$$

Put value in equation (3)

$$ 1=A\sin (\pi \times 0+\pi /4)=A\sin \left( \pi /4 \right) $$

$$ A=\sqrt{2}\,cm $$

Hence, amplitude is $$\sqrt{2}\,cm$$ and phase is $$\dfrac{\pi }{4}$$

A body executing $$SHM$$ has its velocity $$=16$$ cm/s $$=16\times { 10 }^{ -2 }$$ m/s.

$$\begin{array}{l} \left( \begin{array}{l} 1kg \\ ice \\ at-{ 10^{ \circ  } }C \end{array} \right) \xrightarrow [ { Q }_{ 1 } ]{ absorb } \left( \begin{array}{l} 1kg \\ ice \\ at\, { 0^{ \circ  } }C \end{array} \right) \xrightarrow [ { Q }_{ 2 } ]{ absorb } \left( \begin{array}{l} 1kg \\ water \\ at\, { 0^{ \circ  } }C \end{array} \right)  \\ \left( \begin{array}{l} 1kg \\ water \\ at\, { 100^{ \circ  } }C \end{array} \right) \xrightarrow [ { Q }_{ 3 } ]{ released } \left( \begin{array}{l} 1kg \\ water \\ at\, { 0^{ \circ  } } \end{array} \right)  \\ { Q_{ 3 } }=m\rho \Delta \theta =1\times { 10^{ 3 } }\times 1\times 100 \\ =100\times { 10^{ 3 } }cal \\ { Q_{ 1 } }=m\rho \Delta \theta =1\times { 10^{ 3 } }\times 0.5\times 10 \\ =5\times { 10^{ 3 } }cal \\ { Q_{ 2 } }=m\rho =1\times { 10^{ 3 } }\times 80 \\ =80\times { 10^{ 3 } }cal \\ loft\, \, heat=10\times { 10^{ 3 } }-85\times { 10^{ 3 } } \\ =15\times { 10^{ 3 } }cal \\ Now, \\ 15\times { 10^{ 3 } }=2\times { 10^{ 3 } }\times 1\times \Delta \theta  \\ Hence, \\ \theta =7.5\, C \end{array}$$

In the given figure$$:-$$

Let the time taken to travel Ab and BC be $$t_1 $$  and $$t_2$$ respectively.

Let the angular velocity of the system about the point is suspension at any time be $$\omega$$

It's total mechanical energy is equal to its maximum potential energy $$\dfrac{1}{2}kx_{m}^{2}$$, and its potential energy at  $$t = 0$$  is  $$\dfrac{1}{2}kx_{0}^{2}$$  where  $$x_{o}=x_{m}\cos \left ( \pi /5 \right )$$  in this problem. 

From the data of the problem the relative permittivity of the medium varies as 
$$\varepsilon ( x) =\varepsilon_1e^{-(x/l)\ln \dfrac {\varepsilon_1}{\varepsilon_2}}$$
Hence the local velocity of light 
$$v(x) =\dfrac {c}{\sqrt{\varepsilon(x)}}=\dfrac {c}{\sqrt {\varepsilon_1}}e^{\dfrac {x}{2l}\ln \dfrac{\varepsilon_1}{\varepsilon_2}}$$
Thus the required time $$t=\displaystyle \int_0^t{\dfrac{dx}{v(x)}}=\dfrac {\sqrt{\varepsilon_1}}{c}\displaystyle \int_0^l{e^{-\dfrac {x}{2l}\ln \dfrac{\varepsilon_1}{\varepsilon_2}}}$$
$$dx=\dfrac{\sqrt {\varepsilon_1}}{c}\dfrac{-e^{\dfrac{1}{2}\ln \dfrac{\varepsilon_1}{\varepsilon_2}}+1}{\dfrac {1}{2l}\ln \dfrac {\varepsilon_1}{\varepsilon_2}}=\dfrac {2l}{c}\dfrac {\sqrt{\varepsilon_1}-\sqrt{\varepsilon_2}}{\ln \dfrac{\varepsilon_1}{\varepsilon_2}}$$

Thus x = mg/k As the cable breaks the elevator starts falling with acceleration 'g' We shall work in the frame of reference of the elevator Then we have to use a pseudo force mg upward on the block This force will 'balance' the weight Thus the block is subjected to a net force kx by the spring when it is at a distance x from the position of unstretched spring Hence its motion in the elevator is simple harmonic with its mean position corresponding to the unstretched spring Initially the spring is stretched by  = mg/k where the velocity of the block (with respect to the elevator) is zero Thus the amplitude of the resulting simple harmonic motion is mg/k

Volume of the air chamber $$=V$$
Area of cross-section of the neck $$=a$$
Mass of the ball $$=m$$
The pressure inside the chamber is equal to the atmospheric pressure.
Let the ball be depressed by $$x$$ units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.
Decrease in the volume of the air chamber, $$\triangle V=ax$$
Volumetric strain $$=$$ Change in volume/original volume

At equilibrium position deformation of the string is $$x_0$$,

As system accelerate upward a pseudo force will act downward

The given block will perform SHM because floor is frictionless. It does not perform complete oscillations because of the wall. $$T=0.2sec$$, Also the collision time is much less than time period $$T$$. Amplitude of free oscillations if wall was not there is $$\sqrt{2}$$, Which is maximum compression of the spring. the block will make elastic collision with wall because $$\dfrac{1}{\sqrt{2}}<\sqrt{2}.$$

Force is maximum when the body is at extreme position.

$$A$$ is the point where the rod is hinged. Let $$\theta $$ be the angular displacement from mean position. The extension of spring connected at point $$B$$ is 

$$=b\theta$$ ($$\theta$$ is so small that we take it as linear displacement)

So, restoring force at $$B$$ $$={ k }_{ 2 }b\theta $$

Moment of this force about point $$A$$ $$={ k }_{ 2 }b\theta .b={ k }_{ 2 }b^{ 2 }\theta $$($$b$$ is the distance of force from $$A$$)

The contraction of spring connected at point $$c$$ is $$=a\theta$$

So, restoring force is$$={ k }_{ 1 }a\theta $$

And, the moment of force about $$A$$ is $$={ k }_{ 1 }a\theta .a={ k }_{ 1 }a^{ 2 }\theta $$ ($$a$$ is the distance of point $$A$$ of restoring force)

So, total restoring force$$=({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 })\theta $$

Now the total moment of inertia of the system about point $$A$$ is 

$$I$$= Moment of inertiaabout point $$A$$ $$+$$ Moment of inertial of mass $$m$$ about $$A$$

$$=\cfrac { ml^{ 2 } }{ 3 } +ml^{ 2 }$$

So, the angular acceleration, $$\cfrac { { d }^{ 2 }\theta  }{ { dt }^{ 2 } } =-\cfrac { ({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 })\theta  }{ I } =\cfrac { ({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 })\theta  }{ l^{ 2 }(\cfrac { m }{ 3 } +m) } $$

$$\Rightarrow \cfrac { { d }^{ 2 }\theta  }{ { dt }^{ 2 } } +\cfrac { ({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 })\theta  }{ l^{ 2 }(\cfrac { m }{ 3 } +m) } =0$$

So, agular frequency, $$\omega =\sqrt { \cfrac { ({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 }) }{ l^{ 2 }(\cfrac { m }{ 3 } +m) }  } $$

and frequency, $$f=\cfrac { \omega  }{ 2\pi  } =\cfrac { 1 }{ 2\pi  } \sqrt { \cfrac { ({ k }_{ 1 }a^{ 2 }+{ k }_{ 1 }b^{ 2 }) }{ l^{ 2 }(\cfrac { m }{ 3 } +m) }  } $$

$$\tau=k\theta\Rightarrow k=\cfrac{\tau}{\theta}=\cfrac{0.10}{1}\\I=\cfrac{2}{5}MR^2=\cfrac{2}{5}\times 25\times\cfrac{1}{25}=\cfrac{2}{5}kgm^2$$

$$T=365 days=365\times 24\times 3600s$$

$$N_1 + N_2 = mg$$ .........$$(i)$$

(i) Differential equations of a wave is
$$\frac{d^2y}{dt^2} = v^2\frac{d^2y}{dx^2}  $$      ...(1)
Given equation
$$y$$ = 2 sin x cos vt
$$\frac{dy}{dt}$$ = -2v sin x sin vt               ...(2)
Again $$ y = 2$$ sin x cos vt
$$\therefore \ \ \ \frac{dy}{dx} = 2 cos x \ cos vt$$
and $$\frac{d^2y}{dx^2}= -2sin x \ cos vt \ \ \ \ \ ...(3)$$
From equations (2) and (3)
$$\frac{d^2y}{dt^2} = v^2(-2 sin x.cos vt) = v^2.\frac{d^2y}{dx^2}$$
$$or \ \frac{d^2y}{dt^2}=v^2 \frac{d^2y}{dx^2}$$
This is a one-dimensional differential equation.Hence this solution is possible.
$$(ii) \ y= 5sin 2x \ cos vt$$
$$\therefore \ \frac{dy}{dt}=-5v \ sin2x \ sinvt$$
and $$\frac{d^2y}{dt^2}=-5v^2 \ sin2x \ cosvt$$
Again $$y = 5sin\ 2x \ cosvt$$
$$\therefore \ \frac{dy}{dx} = 10cos2x.cosvt$$
and $$\frac{d^2y}{dx^2}=-20sin2x \ cosvt$$
$$\frac{d^2y}{dx^2} = \frac{v^2}{4}(-20 sin2x \ cos vt)$$
$$or \ \frac{d^2y}{dt}= \frac{v^2}{4} \frac{d^2y}{dx^2}$$

Given:
Time period $$ (T) = 4 sec $$
To find:
The time required for the particle to displace $$ \left( \frac { 1 }{ 3 }  \right)rd $$  of the amplitude
The displacement equired of the particle is given by $$ x= A sin \omega t $$ Initially the particle is at the mean position so $$ t=0 $$ and $$ x=0 $$ Let after time $$ t $$ is crosses $$ \left( \frac { 1 }{ 3 }  \right)rd $$ of its amplitude $$ x= A sin \omega t $$
$$ \left( \frac { A }{ 3 }  \right)  = A \quad sin \quad \omega t $$
$$ sin \omega t =\frac{1}{3} $$
By applying the value of $$ sin ^{-1} \left( \frac { 1 }{ 3 }  \right) \omega t =0.3398 $$
$$ \left( \frac { 2 \pi}{ T }  \right) t=0.3398 $$
$$ t= 0.3398 \times \frac{4}{2 \pi} $$
$$ t= 0.216 sec $$
The time required for the particle to displace $$ \left( \frac { 1 }{ 3 }  \right)  rd $$ of the amplitude is $$ 0.216 s $$