Simple Harmonic Motion

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$.

If we substitute these initial conditions into the equation for the displacement of the particle:

$x = A \sin(wt + \delta)$

We have:

$\frac{A}{2} = A \sin(\delta)$

Dividing both sides by $A$ gives us:

$\frac{1}{2} = \sin(\delta)$

The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta = \frac{\pi}{6}$.

$\omega = \sqrt {{K \over m}} \Rightarrow \omega \propto {1 \over {\sqrt m }}$

${{{\omega _2}} \over {{\omega _1}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{1 \over 2}} $

Given

$ x=4(\cos \pi t+\sin \pi t) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The standard equation for amplitude is

$ x=A(\sin \omega t+\phi) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

We have to correct Eq. (i) in format of Eq. (ii) to compare the amplitude.

$x=4[\cos \pi t+\sin \pi t] \times \frac{\sqrt{2}}{\sqrt{2}}$ [multiply and divide it by $\sqrt{2}$ ]

$ \begin{aligned} & x= 4 \sqrt{2}\left[\frac{1}{\sqrt{2}} \cos \pi t+\frac{1}{\sqrt{2}} \sin \pi t\right] \\ & x= 4 \sqrt{2}\left[\cos \pi t \sin \frac{\pi}{4}+\sin \pi t \cos \frac{\pi}{4}\right] \\ & {\left[\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4} \text { and } \cos \frac{\pi}{4}\right] } \\ & x=4 \sqrt{2} \sin \left(\pi t-\frac{\pi}{4}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

because $[\sin (A+B)=[\sin B \cos A +\cos B \sin A] $

Compare Eqs. (iii) and (ii)

$ A=4 \sqrt{2} $

Hence, the amplitude is $4 \sqrt{2}$.

Given,

SHM displacement $x=2 \cos (t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Compare the displacement with standard displacement of SHM

$ x=A \cos (k x+\omega t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Now comparing Eq. (i) and (ii)

$ A=2, K=0, \omega=1 $

where $\omega$ is angular frequency.

$\omega=\frac{2 \pi}{T}$ where $T$ is time period.

$ T=\frac{2 \pi}{\omega} \Rightarrow \frac{2 \pi}{1} s $

So, the time period of particle is $2 \pi$ seconds.

Given force, $F=6.4 \mathrm{~N}$

Elongation in spring $(\Delta x)=0.1 \mathrm{~m}$

time period, $T=\pi / 4$

Time period of oscillation is given by

$ T=2 \pi \sqrt{\frac{m}{K}} $

Since we know, $F=K \Delta x$

$ \begin{array}{cc} & 6.4=K(0.1) \\ \Rightarrow & K=64 \mathrm{~N} / \mathrm{m} \\ \text { Thus, } & T=2 \pi \sqrt{\frac{m}{64}} \Rightarrow \frac{\pi}{4}=\frac{2 \pi \sqrt{m}}{8} \\ \Rightarrow & m=1 \mathrm{~kg} \end{array} $

The effective acceleration of a bob in water is given by,

$ g^{\prime}=g\left(1-\frac{\sigma}{\rho}\right) $

Where, $\sigma$ and $\rho$ are densities of water and bob respectively.

Since, the periods of oscillation of bob in air and water are given as

$ T=2 \pi \sqrt{\frac{l}{g}} \text { and } T=2 \pi \sqrt{\frac{l}{g^{\prime}}} $

$ \begin{aligned} & \text { We have, } \frac{T}{T^{\prime}}=\sqrt{\frac{g^{\prime}}{g}} \\ & \quad \frac{T}{T^{\prime}}=\sqrt{\frac{g\left(1-\frac{\sigma}{\rho}\right)}{g}} \\ & \Rightarrow \quad \frac{T}{T^{\prime}}=\sqrt{1-\frac{\sigma}{\rho}}=\sqrt{1-\frac{1}{\rho}} \quad(\because \sigma=1) \\ & \quad \frac{T}{T^{\prime}}=\frac{1}{\sqrt{2}}=\sqrt{1-\frac{1}{\rho}} \\ & \Rightarrow \quad \frac{1}{2}=1-\frac{1}{\rho} \\ & \Rightarrow \quad \rho=2 \end{aligned} $

Initial Setup:

Spring constant: $k$

Mass: $m$

Initial time period: $1 \text{ s}$

New Setup:

Replaced mass: $4m$

For a spring-block system in simple harmonic motion (SHM), the time period $t$ is given by:

$ t = 2\pi \sqrt{\frac{m}{k}} $

In the initial setup, we have:

$ 1 = 2\pi \sqrt{\frac{m}{k}} $

Rearranging gives:

$ \frac{1}{4\pi^2} = \frac{m}{k} \quad \Rightarrow \quad k = \frac{m}{4\pi^2} $

Now, substituting this new mass into the time period formula:

$ t_2 = 2\pi \sqrt{\frac{4m}{k}} $

Replacing $k$ with the expression found:

$ t_2 = 2\pi \sqrt{\frac{4m}{\frac{m}{4\pi^2}}} = 2\pi \sqrt{\frac{4\times4\pi^2}{1}} = 2\pi \sqrt{16\pi^2} $

This simplifies to:

$ t_2 = 0.5 \text{ s} $

Thus, the clock now runs slow by 0.5 seconds for every second that passes.

Given,

Particle is executing SHM at $x_1=40 \mathrm{ml}$ from mean position.

$ \mathrm{KE}=\frac{1}{3} \mathrm{KE}_{\max } $

max potential energy of SHM$\begin{aligned} & \mathrm{KE}_{\max }=\mathrm{TE}=\frac{1}{2} K A^2 \\ & \mathrm{KE}=\frac{1}{3}\left(\mathrm{KE}_{\max }\right)=\frac{1}{3} \times \frac{1}{2} K A^2 \\ &=\frac{1}{3} \times \frac{1}{2} K A^2 \\ & \frac{1}{2} K\left(A^2-x^2\right)=\frac{1}{6} K A^2 \\ & A^2-x^2=\frac{1}{3} A^2 \\ & { At } x=4 \mathrm{~cm} \\ & A^2-16=\frac{1}{3} A^2 \\ & \therefore A=2 \sqrt{6} \mathrm{~cm}\end{aligned}

$\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|$

$ \Rightarrow {g_{eff}} = g\cos \theta $

$ \Rightarrow T = 2\pi \sqrt {{l \over {{g_{eff}}}}} $

$ = 2\pi = \sqrt {{L \over {g\cos \theta }}} $

$T \propto \sqrt {1/g} $

$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}$

${4 \over 6} = {R \over {R + h}}$

$ \Rightarrow h = R/2$

$ = 3200$ km

Let $x = A\sin \omega t$

$ \Rightarrow v = A\omega \cos \omega t$

$ \Rightarrow v = \, \pm \,\omega \sqrt {{A^2} - {x^2}} $

$ \Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}$

$\Rightarrow$ Ellipse

Both the springs are in parallel combination in both the diagrams so

${T_1} = 2\pi \sqrt {{{3m} \over {2k}}} $

and ${T_2} = 2\pi \sqrt {{m \over {3k}}} $

So, ${{{T_1}} \over {{T_2}}} = {3 \over {\sqrt 2 }}$

$\omega = \pi = \sqrt {{g \over l}} $

So, $l = {g \over {{\pi ^2}}}$

$ \simeq 99.4$ cm (Nearest value)

$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)$

$ = 4\cos (\omega t)$

$y = 4\sin (\omega t)$

$ \Rightarrow {x^2} + {y^2} = {4^2}$

$\Rightarrow$ The particle is moving in a circular motion with radius of 4 m.

$x = \sin \left( {\pi t + {\pi \over 3}} \right)m$

$ \Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)$

$ = \pi \cos \left( {\pi + {\pi \over 3}} \right)$ at $t = 1\,s$

$ = - {\pi \over 2}$ m/s

or $\left| {{{dx} \over {dt}}} \right| = 157$ cm/s

Time taken by the harmonic oscillator to move from mean position to half of amplitude is ${T \over {12}}$

So, ${T \over {12}}$ = 3

T = 36 sec.

${\omega _1}{A_1} = {\omega _2}{A_2}$

$ \Rightarrow {{{A_1}} \over {{A_2}}} = {{{\omega _2}} \over {{\omega _1}}}$

$ = \sqrt {{{{k_2}} \over {{m_2}}}} \times \sqrt {{{{m_1}} \over {{k_1}}}} = \sqrt {{{9k} \over {100}} \times {{50} \over {2k}}} = {3 \over 2}$

Total mechanical energy for a block in SHM is

$ E=\frac{1}{2} m \omega^2 a^2 $

Given, $E=100 \mathrm{~J}, a=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$

So, $\quad 100=\frac{1}{2} m \omega^2 \times\left(5 \times 10^{-2}\right)^2$

$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} m \omega^2=\frac{100}{25 \times 10^{-4}} \\ & \Rightarrow \quad \frac{1}{2} m \omega^2=4 \times 10^4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, at $t=0$, displacement $x$ is given by

$ \begin{aligned} x & =5 \cos \left(\omega t+\frac{\pi}{4}\right) \\ \Rightarrow x(t=0) & =5 \cos \left(\frac{\pi}{4}\right)=\frac{5}{\sqrt{2}} \mathrm{~cm} \end{aligned} $

Now, potential energy at, $x=\frac{5}{\sqrt{2}} \times 10^{-2}$ m is

$ \begin{aligned} U & =\frac{1}{2} m \omega^2 \cdot x^2 \\ & =4 \times 10^4 \times\left(\frac{5}{\sqrt{2}} \times 10^{-2}\right)^2 \\ & =\frac{4 \times 25}{2}=50 \mathrm{~J} \end{aligned} $

Let displacement of particle in SHM is

$ x=A \cos (\omega t+\phi) $

Then, it's velocity is

$ v=\frac{d x}{d t}=-A \omega \sin (\omega t+\phi) $

Now, at $t=2 \mathrm{~s}, x=0$

and $\omega=\frac{2 \pi}{T}$,

where $T=16 \mathrm{~s}$

$ \begin{array}{llrl} & \text { So, } 0= & A \cos \left(\frac{2 \pi}{16} \times 2+\phi\right) \\ \Rightarrow & & \cos \left(\frac{\pi}{4}+\phi\right) & =0 \\ \Rightarrow & & \frac{\pi}{4}+\phi & =\frac{\pi}{2} \\ \Rightarrow & & \phi & =\frac{\pi}{4} \end{array} $

Also, velocity is $4 \mathrm{~m} / \mathrm{s}$ at $t=4 \mathrm{~s}$. So, $v=-A \omega(\sin (\omega t+\phi))$ gives,

$ \begin{array}{ll} \Rightarrow & 4=-A\left(\frac{2 \pi}{16}\right) \cdot \sin \left(\frac{2 \pi}{16} \times 4+\frac{\pi}{4}\right) \\ \Rightarrow & 4=-A \times \frac{2 \pi}{16} \times \sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\ \Rightarrow & 4=-A \times \frac{2 \pi}{16} \times \cos \frac{\pi}{4} . \\ \Rightarrow & 4=\frac{-32 \sqrt{2}}{\pi} \end{array} $

or magnitude of amplitude is $\frac{32 \sqrt{2}}{\pi}$ units.

Given,

$ A(t)=A_0 e^{b t / 2 m} $

where, $b=70 \mathrm{~g} / \mathrm{s}, m=200 \mathrm{~g}$

Mechanical energy ( $E$ ) of the damped oscillation at any instant $t$ is given by

$ \begin{aligned} & & E & =E_0 e^{-b t / m} \\ & \text { Here, } & E & =\frac{E_0}{4}, t=T^{\prime} \\ & \therefore & \frac{E_0}{4} & =E_0 e^{-b T^{\prime} / m} \\ & \Rightarrow & 4 & =e^{b T^{\prime} / m} \\ & \Rightarrow & \log _e 4 & =\frac{b T^{\prime}}{m} \\ & \Rightarrow & 2 \log _e 2 & =\frac{b T^{\prime}}{m} \end{aligned} $

$ \, \begin{aligned}\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad T^{\prime} & =2 \times 2.3026 \log _{10} 2 \times \frac{m}{b} \\ & =2 \times 2.3026 \times 0.3010 \times \frac{200}{70} \\ & =3.96 \mathrm{~s} \simeq 4 \mathrm{~s} \end{aligned} $

Length of simple pendulum, *

$ l=1 \mathrm{~m} $

Mass of bob, $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$

Radius of circular track, $R=100 \mathrm{~m}$

Speed, $v=10 \mathrm{~m} / \mathrm{s}$

The bob of the simple pendulum will experience acceleration due to gravity which provide centripetal acceleration in the circular motion of the car.

Centripetal acceleration, $a_c=v^2 / R$

where, $v$ is the uniform speed of the car and $R$ is the radius of track.

∴ Effective value of acceleration,

$ \begin{aligned} & \begin{aligned} g^{\prime} & =\sqrt{g^2+\left(a_c\right)^2}=\sqrt{g^2+\left(\frac{v^2}{R}\right)^2} \\ \Rightarrow \quad g^{\prime} & =\sqrt{g^2+\frac{v^4}{R^2}} \end{aligned} \\ & \therefore \text { Time period, } T =2 \pi \sqrt{\frac{l}{g^{\prime}}} \\ & =2 \pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}} \\ & =2 \pi \sqrt{\frac{1}{\sqrt{10^2+\frac{10^4}{(100)^2}}}} \\ & =2 \pi \sqrt{\frac{1}{\sqrt{10^2+1}}} \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & T =\frac{2 \pi}{\sqrt{\frac{1}{\sqrt{101}}}}=\frac{2 \pi}{(101)^{1 / 4}} \\ \,\,\,\,\,\,& =\frac{2 \pi}{\alpha^{1 / 4}} \\ & \therefore \alpha =101 \end{array} $

The given situation is shown below.

$ \begin{aligned} &\text { Effective gravitational acceleration }\\ &\begin{aligned} & =\left(g-\frac{q E}{m}\right) \\ \therefore \quad T & =2 \pi \sqrt{\frac{l}{g-\frac{q E}{m}}} \end{aligned} \end{aligned} $

$ \text { The given situation is shown as } $

Let the angle be $\omega t$ when the KE is $75 \%$ of total energy.

Also, the moment of inertia of the bob of pendulum about the point of suspension

$ =I=m l^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, velocity at that instant

$ =l \omega \cos \omega t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, according to the question,

$ \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} I \omega^2 $

where, $\left(\frac{1}{2} I \omega^2\right)$ is the rotational KE of the particle,

$ \begin{aligned} & \frac{1}{2} m v^2=\frac{75}{100} \times \frac{1}{2} \times m l^2 \times \omega^2 \text { [from Eq. (i)] } \\ & \Rightarrow \frac{1}{2} l^2 m \omega^2 \cos ^2 \omega t=\frac{3}{4} \times \frac{1}{2} \times m l^2 \omega^2 \end{aligned} $

[from Eq. (ii)]

$ \begin{aligned} & \Rightarrow \quad \cos ^2 \omega t=3 / 4 \\ & \Rightarrow \quad \cos \omega t= \pm \frac{\sqrt{3}}{2} \Rightarrow \omega t=\frac{\pi}{6} \end{aligned} $

(keeping only positive solution)

$ t=\frac{\pi}{6 \omega}=\frac{\pi T}{6 \times 2 \pi} $

( $\therefore$ Time period, $T=\frac{2 \pi}{\omega}$ )

$ \begin{array}{ll} \Rightarrow & t=\frac{\pi \times 4}{12 \pi} \,\,(given, T=4 \mathrm{~s} )\\ \Rightarrow & t=\frac{1}{3} \mathrm{~s} \end{array} $

Given, instantaneous displacement of particle executing SHM,

$\begin{aligned} x & =A \sin ^2(\omega t-4) \\ \Rightarrow x & =A\left[\frac{1-\cos 2(\omega t-4)}{2}\right] \\ & \left(\text { Since, } \sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right) \end{aligned}$

$\Rightarrow x=\frac{A}{2}[1-\cos (2 \omega t-8)]$

Here, angular velocity of particle

$\begin{aligned} & \omega^{\prime}=2 \omega \\ & \Rightarrow \quad \frac{2 \pi}{T^{\prime}}=2 \times \frac{2 \pi}{T} \Rightarrow T^{\prime}=\frac{T}{2} \\ & =\frac{\frac{2 \pi}{\omega}}{2}=\frac{\pi}{\omega} \\ & \left(\therefore T=\frac{2 \pi}{\omega}\right) \\ \end{aligned}$