Simple Harmonic Motion

Let's denote the amplitude of the simple harmonic motion as A, and the displacement of the particle from the mean position as x. The given condition is that x = A/2.

For a particle in SHM, the potential energy (PE) is given by:

$PE = \frac{1}{2} kx^2$

And the total mechanical energy (E) of the particle remains constant and is given by:

$E = \frac{1}{2} kA^2$

Since the total mechanical energy is the sum of potential energy and kinetic energy (KE), we have:

$E = PE + KE$

Now, we need to find the ratio of potential energy to kinetic energy when x = A/2.

Calculate the potential energy at x = A/2:

$PE = \frac{1}{2} k\left(\frac{A}{2}\right)^2 = \frac{1}{8} kA^2$

Substitute the expression for total mechanical energy:

$KE = E - PE = \frac{1}{2} kA^2 - \frac{1}{8} kA^2 = \frac{3}{8} kA^2$

Now, find the ratio of potential energy to kinetic energy:

$\frac{PE}{KE} = \frac{\frac{1}{8} kA^2}{\frac{3}{8} kA^2} = \frac{1}{3}$

Therefore, the ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude is 1 : 3.

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}

The motion of the particle performing S.H.M. represented by

$\begin{aligned} x & =A \sin \omega t=1 \sin \frac{2 \pi}{T} \cdot t=\sin \frac{2 \pi}{8} t \\ \therefore \quad x & =\sin \left(\frac{\pi t}{4}\right) \end{aligned}$

$\therefore$ Velocity, $v=\frac{d x}{d t}$

$=\frac{d}{d t} \sin \left(\frac{\pi}{4}\right) t=\frac{\pi}{4} \cos \left(\frac{\pi}{4}\right) t$

$\therefore$ Acceleration of the particle.

$\begin{aligned} a & =\frac{d v}{d t}=\frac{d}{d t}\left[\frac{\pi}{4} \cos \left(\frac{\pi}{4}\right) t\right] \\ & =-\frac{\pi}{4} \cdot \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4} t\right) \Rightarrow a=-\frac{\pi^2}{16} \sin \left(\frac{\pi}{4} \cdot t\right) \end{aligned}$

At $t=2 \mathrm{~s}$

$a=-\frac{\pi^2}{16} \cdot \sin \left(\frac{\pi}{4} \cdot 2\right)=-\frac{\pi^2}{16} \mathrm{~ms}^{-2}$

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

$ \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} $