Oscillations and Simple Harmonic Motion

The amplitude, a, at time $t$ is given by $a=a _0\;exp(-\,\alpha t)$

$a _{50}=a _0\;exp(-\alpha\times 50T)=0.80\;a _0$

where $T$ is the period of oscillation

$a _{150}=a _0\;exp(-a\times 150T)$

$=a _0\;(0.8)^3=0.512\,a _0$

Its is a damped oscillation, where amplitude of oscillation at time $t$ is given by $A = a _0e^{-\gamma t}$
where $a _0 = $ initial amplitude of oscillation
$\quad \gamma = $ damping constant
As per question, $\displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma100/v}\quad                    ...(i)$
(where $v$ is the frequency of oscillation)
and $A = a _0e^{-\gamma200/v} \quad                ...(ii)$
From $(i)$; $\quad \displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma\times100/v} \quad            ...(iii)$
Dividing equation $(ii)$ by $(iii)$, we have

$\quad \displaystyle\dfrac{A}{a _0(1/3)} = \displaystyle\dfrac{e^{-\gamma\times200/v}}{e^{-\gamma\times100/v}} = e^{-\gamma\times100/v} = \displaystyle\dfrac{1}{3}$

or $A = a _0\times\displaystyle\dfrac{1}{3}\times\displaystyle\dfrac{1}{3} = \displaystyle\dfrac{1}{9}a _0$

To and fro motion of a particle about mean position is called vibrational motion.

In simple harmonic motion, the average kinetic energy and the average potential energy over one complete period are equal, both being half of the total energy.

Let the equation of $S.H.M$ is:-

Given,

$\begin{array}{l} \sin  \omega t-\cos  \omega t \ =\sqrt { 2 } \left[ { \frac { 1 }{ { \sqrt { 2 }  } } \sin  \omega t-\frac { 1 }{ { \sqrt { 2 }  } } \cos  \omega t } \right]  \ =\sqrt { 2 } \left[ { \sin  \omega t\times \cos  \frac { \pi  }{ 4 } -\cos  \omega t\times \sin  \frac { \pi  }{ 4 }  } \right]  \ =\sqrt { 2 } \sin  \left( { \omega t-\frac { \pi  }{ 4 }  } \right)  \ this\, \, function\, \, represents\, \, SHM\, \, as\, \, it\, \, can\, \, be\, \, written\, \, in\, \, the\, \, form: \ a\sin  \left( { \omega t+\phi  } \right)  \ its\, \, period\, \, is,\, \, \frac { { 2\pi  } }{ \omega  }  \end{array}$

$T=\dfrac{ \pi}{2}$
$V _{max}=1 m/sec$
$\omega =\dfrac{2\pi}{T}$
$V=4  rad/sec$
$V _{max=}A\omega$
$A\times 4=1$
$A=\dfrac {1}{4}$
$a=\omega^{2}x$
$F= m\omega^{2}x$
$x=A   cos   60^o$
$\therefore x=\dfrac {A}{2}$
$F=0.5\times (4)^{2}\times \dfrac {1}{4}\times \dfrac{1}{2}$
$F=1N$

The equation is d^2x/dt^2 + pi^2 * x = 0. Comparing to d^2x/dt^2 + w^2 * x = 0, we get w^2 = pi^2, so w = pi. Since w = 2 * pi * f, then pi = 2 * pi * f, which gives f = 1/2 Hz.

When a particle is dropped from a height $h$ above the centre of tunnel.
$(i)$ It will oscillate, through the earth to a height $h$ on both sides
$(ii)$ The motion of particle is periodic
$(iii)$ The motion of particle will not be $SHM$.

At t=1, the phase of Y1 is pi/2 + phi. The phase of Y2 is 2pi/3 + phi. The difference is 2pi/3 - pi/2 = 4pi/6 - 3pi/6 = pi/6.

${X _1} = A\sin \left( {wt + {\phi _1}} \right)$

At x = A/2, the phase angles are pi/6 and 5pi/6. Since they move in opposite directions, one is at pi/6 (moving away from equilibrium) and the other is at 5pi/6 (moving toward equilibrium). The difference is 5pi/6 - pi/6 = 4pi/6 = 2pi/3.

v1 = dy1/dt = 0.1 * 100pi * cos(100pi*t + pi/3). v2 = dy2/dt = -0.1 * pi * sin(pi*t). This question is garbled regarding the frequencies (100pi vs pi), making a standard phase difference calculation between them invalid.