Physics M.C.Q (1 Marks)

Amplitude $=5 \mathrm{~m}$

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

$T=\frac{2 \pi}{\pi}=2 \mathrm{~s}$

$T=2 \pi \sqrt{\frac{\ell}{g}}$

$T^{\prime}=\frac{x}{2} T$

$2 \pi \sqrt{\frac{\ell}{2 g}}=\frac{x}{2} 2 \pi \sqrt{\frac{\ell}{g}}$

$\frac{1}{\sqrt{2}}=\frac{x}{2} \Rightarrow x=\sqrt{2}$

$\frac{ dx }{ dt }= v = A \omega \cos (\omega t )$

$\frac{ dv }{ dt }= a =-\omega^2 A \sin (\omega t )$

$a =-\left(\frac{2 \pi}{8}\right)^2 \times 1 \sin \left(\frac{2 \pi}{8} \times 2\right)$

$\Rightarrow a =-\frac{\pi^2}{16} \times \sin \left(\frac{\pi}{2}\right)$

$\therefore a =\frac{-\pi^2}{16}\,m / s ^2$

$( n ) 2 \pi \sqrt{\frac{1.21}{ g }}=( n +1) 2 \pi \sqrt{\frac{1}{ g }}$

$( n )(1.1)=( n +1)$

$0.1( n )=1$

$n =10$

No. of oscillation of smaller one

$= n +1$

$=10+1$

$=11$

$(b) \rightarrow (iii)$ $\quad F \propto- x$

$(c) \rightarrow (ii)$ Amplitude is constant

$(d) \rightarrow (i)$ $\quad$ K.E. $+$ P.E. $=$ M.E. $=$ constant

$\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t})=\mathrm{A} \sin (2 \pi \mathrm{nt})$

Now,

Potential energy

$U=\frac{1}{2} k x^{2}=\frac{1}{2} K A^{2} \,\sin ^{2}(2 \pi n t)$

$=\frac{1}{2} k A^{2}\left[\frac{1-\cos (2 \pi(2 n) t)}{2}\right]$

So frequency of potential energy $=2 \mathrm{n}$

$10=\mathrm{K} \times 0.05$

$\mathrm{~K}=\frac{1000}{5}=200$

$\mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2 \pi \sqrt{\frac{2}{200}}$

$=\frac{2 \pi}{10}=\frac{6.28}{10}$

$=.628\, \mathrm{~s}$

$f(t)=f(t+T)$

where $T$ is time period of function

$\sin (\omega t+2 \pi)+\cos (\omega t+2 \pi)$

$=\sin \omega t+\cos \omega t$

$x = A \sin (\omega t +\phi)$ $....(1)$

$\frac{ dx }{ dt }= A \omega \cos (\omega t +\phi)$

acceleration $(a)=\frac{d^{2} x}{d t^{2}}$

$a=-\omega^{2} A \sin (\omega t+\phi)$

$a =\omega^{2} A \sin (\omega t +\phi+\pi)$$.....(2)$

from $(1) \;and\;(2),$ phase difference between displacement and acceleration is $\pi .$

$y=A_{0}+\sqrt{A^{2}+B^{2}} \sin (\omega t+\phi)$

$\mathrm{A}_{0}$ is mean position, and $\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$ is amplitude

$\Rightarrow$ Average velocity $=\frac{\text { Dtsplacement }}{\mathrm{T}}=0$

$\Rightarrow \quad 20=\omega^{2}(5) \Rightarrow \omega=2 \mathrm{rad} \mathrm{s}^{-1}$

$\therefore \quad$ Time period of oscillation, $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \mathrm{s}$

When we cut the spring into ratio of length $1: 2: 3,$ we

get three springs of lengths $\frac{l}{6}, \frac{2 l}{6}$ and $\frac{3 l}{6}$ with force

constant,

$\therefore k_{1}=\frac{k l}{l_{1}}=\frac{k l}{l / 6}=6 k$

${k_{2}=\frac{k l}{l_{2}}=\frac{k l}{2 l / 6}=3 k}$

${k_{3}=\frac{k l}{l_{3}}=\frac{k l}{3 l / 6}=2 k}$

When connected in series,

$\frac{1}{k^{\prime}}=\frac{1}{6 k}+\frac{1}{3 k}+\frac{1}{2 k}=\frac{1+2+3}{6 k}=\frac{1}{k}$

$\therefore \quad \overline{k^{\prime}}=k$

When connected in parallel,

${k^{\prime \prime}=6 k+3 k+2 k=11 k}$

${\frac{k^{\prime}}{k^{\prime \prime}}=\frac{k}{11 k}=\frac{1}{11}}$

The velocity of a particle in simple harmonic motion is given as

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

and magnitude of its acceleration is

$a=\omega^{2} x$

Given $|v|=|a|$

$\therefore \omega \sqrt{A^{2}-x^{2}}-\omega^{2} x$

${\omega x=\sqrt{A^{2}-x^{2}} \text { or } \omega^{2} x^{2}=A^{2}-x^{2}}$

${\omega^{2}=\frac{A^{2}-x^{2}}{x^{2}}=\frac{9-4}{4}=\frac{5}{4}}$

${\omega=\frac{\sqrt{5}}{2}}$

Time period, $T=\frac{2 \pi}{\omega}=2 \pi \cdot \frac{2}{\sqrt{5}}=\frac{4 \pi}{\sqrt{5}} \mathrm{s}$

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

For given spring, $T \propto \sqrt{m}$

$\frac{T_{1}}{T_{2}} =\sqrt{\frac{m_{1}}{m_{2}}}$

$\text { Here, } T_{1} =3 \mathrm{s}, m_{1}=m, T_{2}=5 \mathrm{s}, m_{2}=m+1, m=?$

$ \frac{3}{5}=\sqrt{\frac{m}{m+1}} \text { or } \frac{9}{25}=\frac{m}{m+1}$ 

$25 m=9 m+9 \Rightarrow 16 m=9$

$\therefore m=\frac{9}{16} \mathrm{kg}$

$F = - Kx + {F_0}$$\Rightarrow$  $0 = - Kx + {F_0}$$\Rightarrow$ $x = \frac{{{F_0}}}{K}$

i.e. the particle will oscillate about $x = \frac{{{F_0}}}{K}$

$ = 0.1\sin 2(10\pi \,t + 1.5\pi )$

$ = 0.1\sin (20\pi t + 3.0\pi )$

$\therefore$ Time period, $T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{20\pi }} = \frac{1}{{10}} = 0.1sec$

or $\frac{d^{2} y}{d t^{2}}=\frac{-9}{4} y$

Comparing with $SHM$ equation

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\omega^{2} \mathrm{y}$

$\therefore \omega^{2}=\frac{9}{4}$

$\therefore \omega=\frac{3}{2}$

As this motion is not represented by single harmonic function, hence it is not $SHM.$ As this motion involves sine and cosine functions, hence it is periodic motion.

$x=x_{0}\left[\frac{1-\cos 2 \omega t}{2}\right]=\frac{x_{0}}{2}-\frac{x_{0} \cos 2 \omega t}{2}$

Angular Frequency $=2 \omega$

$\frac{2 \pi}{\mathrm{T}}=2 \omega$

$\mathrm{T}=\frac{\pi}{\omega}$

$=\frac{T}{3}$

$=0.4 \mathrm{s}$

$\because T=18 s$

So distance travelled in $36 \mathrm{sec}=8 \mathrm{A}$

$=8 \times 3=24 \mathrm{cm}$

$\Longrightarrow x=2 \cos \omega t=2 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left[\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]$

On comparing this with equation of $S H M: x=A \sin (\omega t+\phi)$

$\Longrightarrow A=2$           $\omega(\text { Angular frequency })$

$\Longrightarrow T=\frac{2 \pi}{\omega}$

Here, $y=2.5 cm$

$\therefore 2.5=5 \sin \frac{2 \pi t}{6}$

$\Rightarrow \frac{\pi}{6}=\frac{2 \pi t}{6} \Rightarrow t=\frac{1}{2} s$

$\therefore$ The phase $=\frac{2 \pi t}{6}=\frac{2 \pi}{6} \times \frac{1}{2}=\frac{\pi}{6}$

Time taken from mean position to the maximum displacement $ = \frac{1}{4}T = 1\,sec.$

==> $\frac{a}{{\sqrt 2 }} = a\sin \frac{{2\pi }}{T} \cdot t$ 

==> $\sin \frac{{2\pi }}{T}t = \frac{1}{{\sqrt 2 }} = \sin \frac{\pi }{4}$

==> $\frac{{2\pi }}{T}t = \frac{\pi }{4}$

==> $t = \frac{T}{8}$

==> $\frac{A}{2} = A\sin \frac{{2\pi t}}{T}$

==> $t = \frac{T}{{12}}$.

$\frac{a}{2} = a\cos \omega t$

$\Rightarrow \cos \omega t = \frac{1}{2}$

$\Rightarrow \omega t = \frac{\pi }{3}$

$ \Rightarrow \frac{{2\pi t}}{T} = \frac{\pi }{3}$

$\Rightarrow t = \frac{{\frac{\pi }{3} \times T}}{{2\pi }} = \frac{4}{{3 \times 2}} = \frac{2}{3}\sec $

==> $\frac{a}{2} = a\sin \frac{{2\pi t}}{3}$

==> $\frac{1}{2} = \sin \frac{{2\pi t}}{3}$

==> sin$\frac{{2\pi t}}{3} = \sin \frac{\pi }{6}$

==> $\frac{{2\pi t}}{3} = \frac{\pi }{6}$ 

==> $t = \frac{1}{4}\sec $

By solving ${t_1} = 1$sec (For $y = 4 \,cm$) ${t_2} = 3sec$

So time taken by particle in going from 2 cm to extreme position is ${t_2} - {t_1} = 2sec$. 

Hence required ratio will be $\frac{1}{2}$.

Given equation of SHM: $y=a \sin \pi(t+4)$

We will the above equations$:$

hence it will be written as after multiplying by $2$ at $R.H.S$

$y=a \sin 2 \pi\left(\frac{t}{2}+\frac{4}{2}\right)$

$T=2 ; A=5$

$\boldsymbol{x}=\boldsymbol{a} \sin \omega t$

$\Rightarrow x=a \sin \frac{2 \pi t}{T}$

$\Rightarrow x\left(t=\frac{T}{8}\right)=a \sin \left(\frac{2 \pi}{T} \times \frac{T}{8}\right)=a \sin \left(\frac{\pi}{4}\right)=\frac{a}{\sqrt{2}}$

$\Rightarrow \quad 5 \mathrm{cm}=8 \cos \omega t$

$\Rightarrow \quad \omega t=0.9$

$\frac{2 \pi}{1.2} t=0.9$

$\Rightarrow \quad t=\frac{1.08}{2 \pi} \mathrm{s}=0.17 \mathrm{s}$

$\omega=\frac{30 \times 2 \pi}{60}=\pi$

or $\frac{2 \pi}{\mathrm{T}}=\pi$

$\therefore \mathrm{T}=2 \mathrm{sec}$

$x=a \sin (\omega t)$ and $y=a(1-\cos (\omega t))$ where $a$ and $\omega$ are constants.

$\Rightarrow \sin ^2(\omega t)=\frac{x^2}{a^2}$

$\Rightarrow \cos ^2(\omega t)=\left(1-\frac{y}{a}\right)^2$

$\Rightarrow \frac{x^2}{a^2}+\left(1-\frac{y}{a}\right)^2=1$

$\Rightarrow x^2+(y-a)^2=a^2$

The given equation is the equation of circle with centre $(0, a)$, the radius a also the angular velocity is $\omega$ so distance covered $=a c o t$

$x=\sqrt{A^{2}+B}\left[\frac{A}{A^{2}+B^{2}} \sin \omega t+\frac{B}{\sqrt{A^{2}+B^{2}}} \cos \omega t\right]$

$\mathrm{x}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}[\cos \phi \sin \omega \mathrm{t}+\sin \phi \cos \omega \mathrm{t}]$

$\mathrm{x}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}} \sin (\omega \mathrm{t}+\phi)$

amplitude $\sqrt{2}$

$\mathrm{T}=\frac{2 \pi}{\omega}$

$\alpha=60^{\circ}$

$60^{\circ}$ angle travelled in $1\, sec.$

$360^{\circ}$ angle travelled in $1\, sec.$ $\frac{360}{60} \times 1=6$ sec.

$y_{2}=3(\sqrt{2} \sin 3 \pi t+\cos 3 \pi t)$

$=6\left[\frac{\sqrt{3}}{2} \sin 3 \pi t+\frac{1}{2} \cos 3 \pi t\right]$

$6\left[\sin \left(3 \pi t+\frac{\pi}{3}\right)\right]$

$=6 \sin \left(3 \pi t+\frac{\pi}{3}\right)$

ratio of their amplitude is $1 .$

Hence,

Option $A$ is correct answer.

Phase for $2 \,nd$ $\mathrm{SHM}: \phi_{2}=8 \pi \mathrm{t}+\frac{\pi}{4}$

Phase difference : $\Delta \phi=\phi_{1}-\phi_{2}=2 \pi t+\frac{\pi}{12}$

Put $t=0.5 \mathrm{s}$

$\Delta \phi=2 \pi(0.5)+\frac{\pi}{12}=\frac{13 \pi}{12}$

$y_{2}=5 \cos 10 t$

$\Rightarrow \mathrm{y}_{2}=5 \sin \left(10 \mathrm{t}+\frac{\pi}{2}\right)$

phase difference $\frac{\pi}{2}-\phi$

$\Omega=\omega \sqrt{\theta_{0}^{2}-\theta^{2}}$

$\Omega=5 \pi \sqrt{\pi^{2}-\frac{\pi^{2}}{4}} \Omega=5 \pi \sqrt{\pi^{2}-\frac{\pi^{2}}{4}}$

$=\frac{5 \pi^{2} \sqrt{3}}{2}=42.7 \mathrm{rad} / \mathrm{s}$

$\mathrm{x}=\mathrm{A} \sin \left(\omega \mathrm{t}+\frac{5 \pi}{6}\right)=\mathrm{A} \sin \left(\frac{2 \mathrm{n}}{\mathrm{T}} \mathrm{t}+\frac{5 \mathrm{n}}{6}\right)$

$\Delta \phi=120^{\circ}$ between $1 \& 3$

$A_{\text {net }}=A+A=2 A$

$F_{1}=M\left(\frac{2 \pi}{T_{1}}\right)^{2} V ; \quad$ where $\quad M \quad$ is the mass and

$v$ is the velocity and $F_{2}=M\left(\frac{2 \pi}{T_{2}}\right)^{2} V$

$F_{N e t}=F_{1}+F_{2}$

$M\left(\frac{2 \pi}{T}\right)^{2}=M\left(\frac{2 \pi}{T_{1}}\right)^{2}+M\left(\frac{2 \pi}{T_{2}}\right)^{2}$

$\frac{1}{T^{2}}=\frac{1}{T_{1}^{2}}+\frac{1}{T_{2}^{2}}=\frac{T_{1}^{2}+T_{2}^{2}}{T_{1}^{2} T_{2}^{2}}=\frac{T_{1} T_{2}}{\sqrt{T_{1}^{2}+T_{2}^{2}}}$

$\mathrm{x}_{2}=-\mathrm{A} \sin \omega \mathrm{t}$

if they will cross each other

$X_{1}=X_{2}$

for crossing first time

time $=\frac{3 \mathrm{T}}{8}$

$\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+5 \pi / 6)$

$x=A \sin \omega t$

When displacement

$x=\frac{A}{2}$

$\frac{A}{2}=A \sin (\omega t+\phi)$

$\sin ^{-1} \frac{1}{2}=\omega t+\phi$

$\omega t+\phi=30^{\circ} \text { or } 150^{\circ}$

When particles are in opposite direction at one lime phase is $30^{\circ}$ and at the other $150^{\circ}$. So phase difference is $120^{\circ}$.

$T=8 \,s$

$ \omega=\frac{2 \pi}{8}=\frac{\pi}{4}$

$x_1=A \sin \frac{\pi}{4}=\frac{A}{\sqrt{2}}$

$x_2=A \sin \frac{\pi}{4} \times 2-A \sin \frac{\pi}{4}=A-\frac{A}{\sqrt{2}}=\frac{A}{\sqrt{2}}(\sqrt{2}-1)$

$\frac{x_1}{x_2}=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$

Time period is $12 \,s$ from diagram.

$\omega=\frac{2 \pi}{12}=\frac{\pi}{6}$

Amplitude $A=4$

Initial phase is determined by putting known values in the equation.

$2=4 \sin \left(\frac{\pi}{6} t+\phi\right)$

$\sin ^{-1} \frac{1}{2}=\phi[t=0]$

$\frac{\pi}{6}=\phi$

Hence equation is $x=\left(\frac{\pi}{6} t+\frac{\pi}{6}\right)$

$x_1=A \sin \left(\omega t+\phi_1\right)$

$x_2=A \sin \left(\omega t+\phi_2\right)$

$x_1-x_2=A \sin \left(\omega t+\phi_1\right)-A \sin \left(\omega t+\phi_2\right)$

$20=2 \times 20 \sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left[\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right]$

$\frac{1}{2}=\sin \left(\frac{\phi_1-\phi_2}{2}\right) \cdot \cos \left(\omega t+\left(\frac{\phi_1+\phi_2}{2}\right)\right) \text { for maximum value. } \Rightarrow \frac{\phi_1-\phi_2}{2}=\frac{\pi}{6} \Rightarrow \phi_1-\phi_2=\frac{\pi}{3}$

==> $a = \frac{{{v_{\max }} \times T}}{{2\pi }}$

$A = \frac{{1.00 \times {{10}^3} \times (1 \times {{10}^{ - 5}})}}{{2\pi }} = 1.59\,mm$

$v = \omega \sqrt {{a^2} - {x^2}} $

$= \frac{{2\pi }}{T}\sqrt {{A^2} - \frac{{{A^2}}}{4}} $

$= \frac{{2\pi }}{T}\sqrt {\frac{{3{A^2}}}{4}}$

$= \frac{{\pi A\sqrt 3 }}{T}$.

==> ${v_{\max }} = a\omega $

==> $\omega = \frac{{100}}{{10}} = 10\,rad/sec$

Hence $v = \omega \sqrt {{a^2} - {y^2}} $

==>$50 = 10\sqrt {{{(10)}^2} - {y^2}} $

==> $y = 5\sqrt 3 \,cm$

==> $\omega = \frac{{{v_{\max }}}}{a} = \frac{{10}}{4}$ 

Now, $v = \omega \sqrt {{a^2} - {y^2}} $

==> ${v^2} = {\omega ^2}({a^2} - {y^2})$

==> ${y^2} = {a^2} - \frac{{{v^2}}}{{{\omega ^2}}}$ 

$y = \sqrt{{a^2} - \frac{{{v^2}}}{{{\omega ^2}}}}$

$= \sqrt{{4^2} - \frac{5^2}{{({\frac{10}{4}})^2}}}$

$ = 2\sqrt 3 \,cm$

From given equation ${a_1} = 3,\;{a_2} = 4,$ and $\phi = \frac{\pi }{2}$

$a = \sqrt {a_1^2 + a_2^2} $$ = \sqrt {{3^2} + {4^2}} = 5$

==> ${v_{\max }} = a\omega = 5 \times 2 = 10$

$v = \frac{{dx}}{{dt}} = - A\omega \sin \left( {\omega \,t + \frac{\pi }{4}} \right)$ 

For maximum speed, $\sin \,\left( {\omega \,t + \frac{\pi }{4}} \right) = 1$

==> $\omega \,t + \frac{\pi }{4} = \frac{\pi }{2}$ or $\omega \,t = \frac{\pi }{2} - \frac{\pi }{4}$

==> $t = \frac{\pi }{{4\omega }}$  

Velocity is maximum at mean position i.e. $x=0$

$\Rightarrow \sin \left(2 \pi t+\frac{\pi}{3}\right)=0$

$\Rightarrow\left(2 \pi t+\frac{\pi}{3}\right)=0, \pi, 2 \pi, 3 \pi \dots$

if we take $\left(2 \pi t+\frac{\pi}{3}\right)=0,$ time will come negative which is not possible. So we take $\left(2 \pi t+\frac{\pi}{3}\right)=\pi$

$\Rightarrow t=\frac{1}{3} s=0.33 s$

for other solutions, we will get higher values of time. but since it is asking for the first time after $t=0,$ the required solution will be $t=0.33 s$

relation between displacement and velocity is

$\frac{x^{2}}{A^{2}}+\frac{v^{2}}{(A \omega)^{2}}=1$

$\therefore A \omega=2$ and $A=4$

$\omega=\frac{1}{2} \quad \Rightarrow \mathrm{T}=4 \pi$

Comparing with $\frac{\mathrm{x}^{2}}{\mathrm{A}^{2}}+\frac{\mathrm{v}^{2}}{(\mathrm{A} \omega)^{2}}=1$

$A=5, A \omega=\frac{5}{2} \quad \therefore \omega=\frac{1}{2} \Rightarrow T=4 \pi$

$\mathrm{v}=\mathrm{A} \omega \cos \omega \mathrm{t} ; \mathrm{a}=\mathrm{A} \omega^{2} \sin \omega \mathrm{t}$

$\Rightarrow \frac{v^{2}}{A^{2} \omega^{2}}+\frac{a^{2}}{A^{2} \omega^{4}}=1$

$(ii)$ and $(iii)$ are correct

$\mathrm{V}_{0}=\mathrm{O} \Rightarrow \mathrm{V}_{\mathrm{I}}=2 \mathrm{V}_{\mathrm{M}}$

$A_{I}=4 \mathrm{cm}$

$\mathrm{V}_{\max }=\mathrm{A} \omega=3 \times \frac{200}{60}=10 \mathrm{ms}^{-1}$

$=-A \omega \sin \left(\omega t+\frac{\pi}{4}\right)$

For maximum speed,

$\sin \left(\omega t+\frac{\pi}{4}\right)=1 \Rightarrow \omega t+\frac{\pi}{4}=\frac{\pi}{2}$

or $\omega t=\frac{\pi}{2}-\frac{\pi}{4} \Rightarrow t=\frac{\pi}{4 \omega}$

$\operatorname{acc}^{\mathrm{n}} \mathrm{f}=\frac{\omega^{2} \mathrm{a}}{2}$

$\therefore \mathrm{V}=\mathrm{f}$

$\frac{\omega^{2} a}{2}=\frac{\sqrt{3}}{2} a \omega$

$2 \pi n=\sqrt{3}$

$\mathrm{n}=\frac{\sqrt{3}}{2 \pi}$

Average velocity, $v_{\text {avg }}=\frac{\text { displacement }}{\text { total time }}$

Given, $y=7 \sin (\pi t) m , t=0$ to $0.5$

$y$ at $t=0, x_1=7 \sin \left(0^{\circ}\right)=0$

$y$ at $t=0.5$,

$x_2=7 \sin (\pi / 2)=7$

$\therefore V_{\text {avg }}=\frac{7-0}{0.5-0}=14$

$v_{\text {avg }}=14 m / s$

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

$v_1=3 \,m / s x_1=4 \,m$

$v_2=4 \,m / s x_2=3 \,m$

$3=\omega \sqrt{A^2-4^2} \ldots(i)$

$4=\omega \sqrt{A^2-3^2} \ldots(ii)$

Solving $(i)$ and $(ii)$, we get

$A=5 \,m$ and $\omega=1 \,rad / s$

$A=10 \,cm \quad A \omega=0.4 \,m / s$

$=0.1 \,m$

$\therefore \omega =4 \,rad / s$

$T=\frac{2 \pi}{4}=\frac{\pi}{2} s$

$m=1 \times 10^{-20} \,kg$

$T=1 \times 10^{-5} \,kg$

Maximum speed $=A \omega=1 \times 10^3 \,m / s \quad \dots (i)$

$\omega=\frac{2 \pi}{T}=2 \pi \times 10^5 \,rad / s$

Putting value of $\omega$ in $(i)$

$A \times 2 \pi \times 10^5=1 \times 10^3$

$A=\frac{1}{2 \pi \times 10^2}$

$A=1.59 \,mm$

Phase at mean position $=0$

Phase at mid point

$\frac{A}{2}=A \sin \phi$

$\phi=\frac{\pi}{6}$

Time it takes to travel a phase difference of $\phi$

$t=\frac{2 \pi}{\omega} \times \frac{\phi}{2 \pi}$

or $t=\frac{\phi}{\omega}$

or $t=\frac{\pi}{6 \omega}$

Average speed $=\frac{\text { Total distance }}{\text { Time taken }}$

$=\frac{A / 2}{\pi / 6 \omega}$

$=\frac{3 A \omega}{\pi}$

$v^2+a x^2=b$

$v^2=b-a x^2$

$v^2=a\left(\frac{b}{a}-x^2\right)$

Comparing it to equation

$v^2=\omega^2\left(A^2-x^2\right)$

$\omega=\sqrt{a}$

$f=\frac{\omega}{2 \pi}=\frac{\sqrt{a}}{2 \pi}$

$x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right]$

At $t=\frac{1}{6} s$

$x=10 \cos \left[\frac{\pi}{2}+\frac{\pi}{3}\right]$

$x=-10 \sin \frac{\pi}{3}$

$x=-5 \sqrt{3}$

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

$v=2 \pi \sqrt{100-75}$

$v=10 \pi$

$v=31.4 \,cm / s$

$ = 0.01 \times 4 \times {(\pi )^2} \times {(60)^2} = 144{\pi ^2}\,m/\sec $

$= \frac{{a \times 4{\pi ^2}}}{{{T^2}}}$

$ = \frac{{1 \times 4 \times {{(3.14)}^2}}}{{0.2 \times 0.2}}$

${F_{\max }} = m \times {A_{\max }}$

$= \frac{{0.1 \times 4 \times {{(3.14)}^2}}}{{0.2 \times 0.2}}$

$= 98.596\;N$

Maximum acceleration $ = {\omega ^2}a = 24$ 

$ \Rightarrow a = \frac{{{{(a\omega )}^2}}}{{{\omega ^2}a}} = \frac{{16 \times 16}}{{24}} = \frac{{32}}{3}m$

Now given, ${\omega ^2}x = \omega \sqrt {{A^2} - {x^2}} $

==> ${\omega ^2}.1 = \omega \sqrt {{2^2} - {1^2}} $

==> $\omega = \sqrt 3 $ 

$T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{\sqrt 3 }}$

From the given value $\frac{{|A|}}{x} = {\omega ^2} = 4$ ==> $\omega = 2.$

Also $\omega = \frac{{2\pi }}{T} \Rightarrow 2 = \frac{{2\pi }}{T}\, \Rightarrow T = \pi \,sec$

$v_{\max }=\mathrm{A} \omega$

Double $\omega$; half the amplitude

$\mathrm{A}=\frac{\mathrm{g}}{\omega^{2}}=\frac{\mathrm{g}}{\left(\frac{2 \pi}{\mathrm{T}}\right)^{2}}=\frac{\mathrm{gT}^{2}}{4 \pi^{2}}$

$A=\frac{10 \times 4}{4 \pi^{2}}=\frac{10}{\pi^{2}} \mathrm{m}$

so graph of $a$ vs $x$ is a straight line having slope $-\omega^{2}$

$\therefore \quad A=\frac{g}{\omega^{2}}=\frac{g}{\left(\frac{2 \pi}{T}\right)^{2}}=\frac{g T^{2}}{4 \pi^{2}}$

$\mathrm{A}=\frac{10 \times 4}{4 \pi^{2}}=\frac{10}{\pi^{2}} \mathrm{m}$

$v = 0$ 

$\Rightarrow \frac{a}{x}=\omega^{2}=\tan 37^{\circ}=\frac{3}{4}$

$\Rightarrow \frac{4 \pi^{2}}{\mathrm{T}^{2}}=\frac{3}{4} \Rightarrow \mathrm{T}=\frac{4 \pi}{\sqrt{3}}$ $s$

$\mathrm{V}_{\max }=\beta \Rightarrow \omega \mathrm{A}=\beta$          $...(2)$

Dividing

$\omega=\frac{\alpha}{\beta} \Rightarrow n=\frac{\alpha}{2 \pi \beta}$

$a=-\omega^{2} x=0.4 \mathrm{m} / \mathrm{s}^{2}$

$\omega^{2}=4 \quad \Rightarrow \quad \omega=2$

$\omega A=0.2$

$2 \times \mathrm{A}=0.2$

$\mathrm{A}=0.1 \mathrm{m}$

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{d} \overrightarrow{\mathrm{r}}}{\mathrm{dt}}=(\cos t \hat{i}-\sin t \hat{j})$

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}=(-\sin t \hat{\mathrm{i}}-\cos t \hat{\mathrm{j}})$

According to question

$\overrightarrow{r} \cdot \overrightarrow{a}=0$

$\Rightarrow-\sin ^{2} t-\cos ^{2} t=0$

$\Rightarrow 1=0$ which is not possible

$4 \frac{d^2 x}{d t^2}+320 x=0$

$4 a=-320 x$

$a=-80 x$

Since $a=-\omega^2 x$ in $S.H.M.$

$80=\omega^2$

$\sqrt{16 \times 5}=\omega$

or $\omega=4 \sqrt{5}$

$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{4 \sqrt{3}}=\frac{\pi}{2 \sqrt{5}} s$

Period $(T)=2 \,s$

$\omega=\frac{2 \pi}{2}=\pi \,rad / s$

When block just represent from a piston, maximum acceleration must be equal to $g$.

$g=-\omega^2 x$

Acceleration is maximum when $x=A$

$g=-\omega^2 A$

or $A=\frac{9.8}{\pi^2}$

Maximum velocity $=A \omega$

$=\frac{9.8}{\pi^2} \times \pi$

$=\frac{9.8}{\pi} \,m / s$

$=3.119 \,m / s =3.12 \,m / s$

$x=5 \cos (2 \pi t+\pi / 4)$

$t=\frac{3}{2} s$

$x=5 \cos (3 \pi+\pi / 4)$

$x=5 \cos \left(\frac{13 \pi}{4}\right)$

$x=-5 \cos \frac{\pi}{4}$

$x=-\frac{5}{\sqrt{2}}$

Acceleration

$a=-\omega^2 x$

$a=-4 \pi^2 x-\frac{5}{\sqrt{2}}$

$a=139.56 \,m / s ^2$

$A=2 \,cm =2 \times 10^{-2}$

$a=-\omega^2 x$

and $v=\omega \sqrt{A^2-x^2}$

$\omega^2 \times 1 \times 10^{-2}=\omega \sqrt{(4-1) 10^{-4}}$

$\omega \times 1 \times 10^{-2}=\sqrt{3} \times 10^{-2}$

$\omega=\sqrt{3}$

$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{3}}$

Maximum acceleration of the system $\left(a_{\max }\right)=-\omega^2 A$

For a block to escape the board the acceleration must be equal to 9 at the top-most point.

$g=\omega^2 A$

$\omega=\sqrt{\frac{g}{A}}$

Time period $=\frac{2 \pi}{\omega}=\sqrt{\frac{A}{g}}$

==>  $x = \frac{a}{\sqrt{2}} =  \frac{a}{\sqrt{2}} = 2\sqrt{2}\,cm$

Potential energy $U = \frac{1}{2}m{\omega ^2}({a^2} - {y^2}) $

$= E - \frac{1}{2}m{\omega ^2}{y^2}$ 

When $y = \frac{a}{2}$

$⇒$ $U = E - \frac{1}{2}m{\omega ^2}\left( {\frac{{{a^2}}}{4}} \right)$

$= E - \frac{E}{4} = \frac{{3E}}{4}$

So $\frac{{2.5}}{E} = \frac{{{{\left( {\frac{a}{2}} \right)}^2}}}{{{a^2}}}$

==> $E = 10J$

==> $\frac{U}{{80}} = \frac{{{{\left( {\frac{3}{4}a} \right)}^2}}}{{{a^2}}} = \frac{9}{{16}}$

==> $U = 45\,J$

$\Rightarrow {v_{\max }} = \sqrt {\frac{{2\,{K_{\max }}}}{m}} $ $ = \sqrt {\frac{{2 \times 16}}{{0.32}}} $

$ = \sqrt {100} $ 

$ = 10\,m/s.$

$ = 4[\sin 3\omega \,t]$ (By using $\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta )$

$\therefore $ maximum acceleration ${A_{\max }} = {(3\omega )^2} \times 4 = 36{\omega ^2}$

Similarly for zero acceleration, velocity is maximum so kinetic energy is also maximum.

Putting $E = K + U$ we obtain,

$A = \frac{1}{{2\pi \left( {\frac{{25}}{\pi }} \right)}}\sqrt {\frac{{2 \times (0.5 + 0.4)}}{{0.2}}} $ $ \Rightarrow A = 0.06\,m$

$ \Rightarrow \frac{1}{2}m{\omega ^2}({a^2} - {x^2}) = \frac{1}{2}m{\omega ^2}{x^2}$

$ \Rightarrow {a^2} - {x^2} = {x^2}$$ \Rightarrow {x^2} = \frac{{{a^2}}}{2}$$ \Rightarrow \frac{x}{a} = \frac{1}{{\sqrt 2 }}.$

==> $\frac{{{n_1}}}{{{n_2}}} = \frac{1}{{\sqrt 2 }}$==> ${n_2} = \sqrt 2 \,{n_1}$ ==> ${n_2} > {n_1}$

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

==>$\frac{{a_1^2}}{{a_2^2}} = \frac{{{m_2}n_2^2}}{{{m_1}n_1^2}}$ ( $E$ is same)

Given ${n_2} > {n_1}$ and ${m_1} = {m_2}$ ==> ${a_1} > {a_2}$

$2 \pi^{2} f^{2} A^{2}(1+\cos 2 \omega t)$

The average potential energy is equal to average kinetic energy is equal to average of $m v^{2} / 2=\frac{1}{2} m \times 2 \pi^{2} f^{2} A^{2}(1+\cos 2 \omega t)-(1)$

$\Rightarrow m \pi^{2} f^{2} A^{2}$

From $( 1 ),$ angular frequency of oscillation of kinetic energy is $2$ $\omega$ i.e., frequency of oscillation of kinetic energy is $2 f$

$\mathrm{PE}=\mathrm{TE} \propto \mathrm{q} \mathrm{A}^{2}$

$\mathrm{PA}^{2} / \omega^{2}=\mathrm{q} \mathrm{A}^{2}$

$\mathrm{T}=2 \pi \sqrt{\frac{p}{q}}$

$\frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} m \omega^{2} A^{2}=m \omega^{2} A^{2}$

Let $A^{\prime}$ be the new amplitude. (apply energy conservation law)

$\frac{1}{2} m_{\omega}^{2}\left(A^{\prime}\right)=m_{\omega}^{2} A^{2}$

${A}^{\prime}=\sqrt{2} A$

that $A^{2}-x^{2}=\frac{x^{2}}{n} \quad$ or $x^{2}\left(1-\frac{1}{n}\right)=A^{2}$

$x=\sqrt{\frac{n}{n+1}} A$

$A=2$

$\frac{1}{2} \mathrm{m} \omega^{2}\left(\mathrm{A}^{2}-\mathrm{y}^{2}\right)=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{y}^{2}$

$y=\frac{A}{\sqrt{2}}$

$\mathrm{U}_{0}=2 \mathrm{J}(\text { given })$ and $\frac{1}{4} \mathrm{KA}^{2}=4 \mathrm{J}$

So total energy $=8+2=10 \mathrm{J}$

$8 \times 10^{4} x^{2}$ erg. $=\frac{1}{2} \omega^{2} x^{2}$

$\omega=400$

total energy $=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2}$

$\therefore A=10 \mathrm{cm}$

$\frac{1}{2} \mathrm{m} \omega^{2}\left(\mathrm{A}^{2}-\mathrm{y}^{2}\right)=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{y}^{2}$

$\Rightarrow \mathrm{y}=\frac{\mathrm{A}}{\sqrt{2}}$

$\mathrm{T}=\frac{1}{2} \mathrm{K}\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)$

$\frac{V}{T}=\frac{x^{2}}{A^{2}-x^{2}}$

According to Hooke's law

$\therefore \quad \mathrm{F}_{\mathrm{p}}=-k_{\mathrm{p}} x_{\mathrm{p}}$

$F_{\mathrm{Q}}=-k_{\mathrm{Q}} x_{\mathrm{Q}} \Rightarrow \frac{F_{p}}{F_{Q}}=\frac{k_{p}}{k_{Q}} \frac{x_{p}}{x_{Q}}$

$F_{\mathrm{p}}=F_{\mathrm{Q}}[\text { Given }]$

$\therefore \frac{x_{p}}{x_{Q}}=\frac{k_{Q}}{k_{p}}$                    $...(i)$

Energy stored in a spring is $\mathrm{U}=\frac{1}{2} k x^{2}$

$\therefore \quad \frac{U_{p}}{U_{Q}}=\frac{k_{p} x_{p}^{2}}{k_{Q} x_{Q}^{2}}=\frac{k_{p}}{k_{Q}} \times \frac{k_{Q}^{2}}{k_{p}^{2}}=\frac{1}{2}\left[\because k_{Q}=\frac{k_{p}}{2}\right]$

$\Rightarrow U_{p}=\frac{U_{Q}}{2}=\frac{E}{2} \quad\left[\therefore U_{Q}=E\right]$

Mass $=4 \,kg$

Maximum $P.E.$ $=\frac{1}{2} k A^2$

$1=\frac{1}{2} \times k \times(0.2)^2$

$\frac{2}{0.04}=k$

$k=50 \,N / m$

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

$T=2 \pi \sqrt{\frac{4}{50}}$

$=\frac{2 \sqrt{2} \pi}{5} \,s$

Energy $=90 \,J \quad$ Amplitude $=6 \,cm$

Maximum energy $=\frac{1}{2} m A^2 \omega^2=90$

$\text { or } m \omega^2=\frac{180}{36 \times 10^{-4}}$

$\therefore m \omega^2=30 \times 10^2$

When particle is stopped the point where it is stopped is the new amplitude but angular velocity will remain same.

$E=\frac{1}{2} m A_2^2 \omega^2$

$\text { or } E=3000 A_2^2$

$A_2=4 \times 10^{-2}$

$A_2=3000$

$U=\frac{1}{2} k x^2$

$x^2=\frac{2 U}{k}$

or $x \propto \frac{1}{k}$ (Since $U$ is constant)

Also $T=2 \pi \sqrt{\frac{m}{k}}$

or $T \propto \frac{1}{\sqrt{k}}$

Therefore $x \propto T$

Hence the oscillation with maximum $x$ will have the maximum time period.

$T=8 \,s$

Maximum value of potential energy is reached two times per oscillation which is $\frac{T}{4}$ time away from mean position which has minimum value at position.

Energy $=E_0$

After a phase shift of $\frac{\pi}{3}$

$E=E_0 \cos ^2 \frac{\pi}{3}$

$\frac{P^2}{2 m}=\frac{E_0 3}{4}$

$P=\sqrt{\frac{3 m E}{2}}$

In upward moving lift $T' = 2\pi \sqrt {\frac{l}{{(g + a)}}} $ ($a = $Acceleration of lift)

$ \Rightarrow \frac{{T'}}{T} = \sqrt {\frac{g}{{g + a}}} = \sqrt {\frac{g}{{\left( {g + \frac{g}{4}} \right)}}} = \sqrt {\frac{4}{5}} $

$ \Rightarrow T' = \frac{{2T}}{{\sqrt 5 }}$

==> $\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{{0.02}}{2} = 0.01$

==>$\Delta T = 0.01\,T$

Loss of time per day $ = 0.01 \times 24 \times 60 \times 60$$ = 864\sec $

$\therefore {g_{eff}} = g\,\left( {\frac{{\rho - 1}}{\rho }} \right)$

$\frac{{T'}}{T} = \sqrt {\frac{g}{{{g_{eff}}}}} $

$\frac{{T'}}{T} = \sqrt {\frac{\rho}{{\rho -1}}} $

$ = g' = g\,\left( {1 - \frac{\sigma }{\rho }} \right)$ where $\sigma$ and $\rho$ are the density of water and the bob respectively. Since the period of oscillation of the bob in air and water are given as $T = 2\pi \sqrt {\frac{l}{g}} $ and $T' = 2\pi \sqrt {\frac{l}{{g'}}} $

$\therefore $ $\frac{T}{{T'}} = \sqrt {\frac{{g'}}{g}} = \sqrt {\frac{{g(1 - \sigma /\rho )}}{g}} $$ = \sqrt {1 - \frac{\sigma }{\rho }} = \sqrt {1 - \frac{1}{\rho }} $

Putting $\frac{T}{{T'}} = \frac{1}{{\sqrt 2 }}$. We obtain, $\frac{1}{2} = 1 - \frac{1}{\rho }$

$ \Rightarrow \rho = 2$

$\omega=\sqrt{\frac{g}{\alpha}}$

$\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\alpha}} \quad \text { or } \quad \mathrm{T}=2 \pi \sqrt{\frac{\alpha}{\mathrm{g}}}$

then $\mathrm{Ah} \rho=\mathrm{A} \ell \rho_{0}$

$\Rightarrow \mathrm{h}=\frac{\rho_{0} \ell}{\rho}$

$\therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{h}}{\mathrm{g}}} \Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\rho_{0} \ell}{\rho \mathrm{g}}}$

$w=\sqrt{\frac{g}{d}}=2 \pi t$

$t=\frac{1}{2 \pi} \sqrt{\frac{g}{d}}$

Radius of the disc, $r=15 \,cm =0.15\, m$

The torsional oscillations of the disc has a time period, $T=1.5\, s$

The moment of inertia of the disc is:

${I}=\frac{1}{2} m r^{2}$

$={2} \times(10) \times(0.15)^{2}$

$=0.1125 \,kg\, m ^{2}$

$T=2 \pi \sqrt{\frac{I}{\alpha}} a$

Time period,

is the torsional constant.

$\alpha=\frac{4 \pi^{2} I}{T^{2}}$

$=\frac{4 \times(\pi)^{2} \times 0.1125}{(1.5)^{2}}$

$=1.972\, Nm / rad$

Hence, the torsional spring constant of the wire is $1.972\, Nm$ $rad$ $^{-1}$.

The time period of the disc is $2 \pi \sqrt{(3 r / 2 g )}$

We know that the time period of an object,

$T =2 \pi \sqrt{(I / mgL )}$

where,

$I=$ moment of inertia from the suspended point

$L =$ distance of its centre from suspended point $= r$

we know that, the moment of inertia of disc about its centre $= mr ^2 / 2$

using parallel axis theoram the moment of inertia from a point in its periphery,

$I = mr ^2+ mr ^2 / 2=3 mr ^2 / 2$

putting the values in the above equation we get,

$2 \pi \sqrt{\left(3 mr ^2 / 2 mgr \right)}$

$=2 \pi \sqrt{(3 r / 2 g )}$

therefore, the time period of the disc is $2 \pi \sqrt{(3 r / 2 g )}$

This is the case of a physical pendulum.

$T=2 \pi \sqrt{\frac{I_{\text {com }}}{m g L_{\text {com }}}}$

$L_{\text {com }}=\frac{L}{2} I_{ com }=\frac{m L^2}{3}$

$T=2 \pi \sqrt{\frac{2 l}{3 g}}$

It is the case of a physical pendulum.

$T=2 \pi \sqrt{\frac{I_{\text {c.o.m. }}}{m g L_{\text {com }}}}$

$I_{\text {com }}=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2$

$L_{\text {com }}=R$

$T=2 \pi \sqrt{\frac{3 R}{2 g}}$

$U(x)=\alpha+2 \beta x^2$

$F=-\frac{d U(x)}{d x}$

$F=-4 \beta x$

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

$T=2 \pi \sqrt{\frac{m}{4 \beta}} \quad \cos [k=\beta]$

$T=\pi \sqrt{\frac{m}{\beta}}$

When bob is inserted, in liquid effective of is reduced because of the force of upthrust.

$\rho_s$ is density of solid.

$\rho_L$ is density of liquid.

$V$ is volume of solid

and $T=2 \pi \sqrt{\frac{1}{g}}$

$T_{\text {new }}=2 \pi \sqrt{\frac{1}{g_{\text {new }}}}$

$T_{\text {new }}=2 \pi \sqrt{\frac{1}{g-\frac{\rho_L}{\rho_S} g}}$

$T_{\text {new }}=2 \pi \sqrt{\frac{1 \times 12}{g \times 11}}$

or $T_{\text {new }}=\sqrt{\frac{12}{11}} T$

Also spring constant $(k) \propto \frac{1}{{{\rm{Length (}}L)}}$, when the spring is half in length, then $k$ becomes twice. 

$\therefore T' = 2\pi \sqrt {\frac{m}{{2k}}} $

$\Rightarrow \frac{{T'}}{T} = \frac{1}{{\sqrt 2 }} $

$\Rightarrow T' = \frac{T}{{\sqrt 2 }}$

==> $\frac{{{n_S}}}{{{n_P}}} = \sqrt {\frac{{{k_S}}}{{{k_P}}}} $

$ \Rightarrow \frac{{{n_s}}}{{{n_p}}} = \sqrt {\frac{{\left( {\frac{k}{2}} \right)}}{{2k}}} = \frac{1}{2}$

==> $m \propto {T^2}$

==> $\frac{{{m_2}}}{{{m_1}}} = \frac{{T_2^2}}{{T_1^2}}$

$ \Rightarrow \frac{{M + m}}{M} = {\left( {\frac{{\frac{5}{4}T}}{T}} \right)^2}$
$ \Rightarrow \frac{m}{M} = \frac{9}{{16}}$

==>$\frac{n}{{n'}} = \sqrt {\frac{k}{m} \times \frac{{m'}}{{K'}}} $

$ = \sqrt {\frac{k}{m} \times \frac{{2m}}{{2K}}} = 1$

==> $n' = n$

$\frac{f}{{f'}} = \sqrt 2 $

==> $f' = \frac{f}{{\sqrt 2 }}$

==> $x = - 0.16\cos (\pi \,t)$       [As $a = -0.16$ meter, $T = 2\, sec$]

$ \Rightarrow \frac{{{F_1}}}{{{F_2}}} = \frac{{{k_1}{x_1}}}{{{k_2}{x_2}}} = \frac{{{k_1}}}{{{k_2}}} \times \sqrt {\frac{{{k_2}}}{{{k_1}}}} = \sqrt {\frac{{{k_1}}}{{{k_2}}}} $

==> ${T_1}:{T_2}:{T_3} = \frac{1}{{\sqrt k }}:\frac{1}{{\sqrt {k/2} }}:\frac{1}{{\sqrt {2k} }} = 1:\sqrt 2 :\frac{1}{{\sqrt 2 }}$

==> $\frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{K_1}}}{{{K_2}}}} $$ = \sqrt {\frac{k}{{4k}}} = \frac{1}{2}$

==> ${T_2} = \frac{{{T_1}}}{2}$

$ \Rightarrow 10g = k \times 0.25$

$\Rightarrow k = \frac{{10g}}{{0.25}} = 98 \times 4$

Now $T = 2\pi \sqrt {\frac{m}{k}}$

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

$ \Rightarrow m = \frac{{{\pi ^2}}}{{100}} \times \frac{1}{{4{\pi ^2}}} \times 98 \times 4 = 0.98\;kg$

==> $mg = kx$

==> $m \propto kx$
Hence $\frac{{{m_1}}}{{{m_2}}} = \frac{{{k_1}}}{{{k_2}}} \times \frac{{{x_1}}}{{{x_2}}}$

==> $\frac{4}{6} = \frac{k}{{k/2}} \times \frac{1}{{{x_2}}}$
$ \Rightarrow {x_2} = 3\,cm$.

$ \Rightarrow mg = kx$

$ \Rightarrow k = \frac{{mg}}{x} = \frac{{1 \times 10}}{{5 \times {{10}^{ - 2}}}} = 200\frac{N}{m}$
Further, the angular frequency of oscillation of $2\, kg$ mass is $\omega = \sqrt {\frac{k}{M}} = \sqrt {\frac{{200}}{2}} = 10\,rad/sec$
Hence, ${v_{\max }} = a\omega = (10 \times {10^{ - 2}}) \times 10 = 1\,m/s$

==>${y^2} = {a^2} - \frac{{{u^2}}}{{{\omega ^2}}}$$ = {(0.5)^2} - \frac{{{{(0.4)}^2}}}{{{1^2}}}$

==> ${y^2} = 0.9$

 $y = 0.3\,m$

$m{\omega ^2}a = m \times {(2\pi f)^2}a = 60 \times {\left( {2\pi \times \frac{2}{\pi }} \right)^2} \times 0.1\,N$

$ = 60 \times 16 \times 0.1 = 96N = \frac{{96}}{{9.8}} \approx 10\,kgf$ and this force is towards mean position.

The reaction of the force on the platform away from the mean position. It reduces the weight of man on upper extreme i.e. net weight $= (60 -10)\, kgf.$

This force adds to the weight at lower extreme position i.e. net weight becomes $= (60 + 10)\, kgf.$

Therefore, the reading the weight recorded by spring balance fluctuates between $50\, kgf$ and $70\, kgf.$

$6.4 = k(0.1)$ or $k = 64N/m$

$T = 2\pi \sqrt {\frac{m}{k}} $ ==> $\frac{\pi }{4} = 2\pi \sqrt {\frac{m}{{64}}} $; $\frac{m}{{64}} = {\left( {\frac{1}{8}} \right)^2}$; $m = 1kg$

Equivalent spring constants of value $k$ and $3k$ are in parallel and their net value of spring constant of all the four springs is $k + 3k = 4k$

$\therefore $ Frequency of mass is $n = \frac{1}{{2\pi }}\sqrt {\frac{{4k}}{M}} $

$\frac{1}{\mathrm{k}_{\mathrm{eq}}}=\frac{1}{\mathrm{k}}+\frac{1}{\mathrm{k}}=\frac{2}{\mathrm{k}}$

$\therefore \quad k_{e q}=\frac{k}{2}$

$\therefore \quad \mathrm{T}_{1}=2 \pi \sqrt{\frac{2 \mathrm{m}}{\mathrm{k}}}$

$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}^{\prime}}}(\text { in parallel })$

But $k^{\prime}=k+k=2 k$

$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$

$\therefore \frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}=\frac{2 \pi \sqrt{\frac{2 \mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}}=2$

$T_{p}=2 \pi \sqrt{\frac{\ell}{g}} ; \quad T_{p} \propto g^{-1 / 2}$

$\therefore T_{p}$ decreases

at equilibrium position

Max. compression $=x-x_{e}$

$\mathrm{K}\left(\mathrm{x}-\mathrm{x}_{\mathrm{e}}\right)=\mathrm{M}_{2} \mathrm{g}$

Now $\quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

or $\frac{\pi}{4}=2 \pi \sqrt{\frac{\mathrm{m}}{64}}$

$\therefore m=1 \mathrm{kg}$

$\frac{3=2 \pi \sqrt{\frac{0.9}{\mathrm{k}}}}{\mathrm{T}=2 \pi \sqrt{\frac{0.4}{\mathrm{k}}}}$

$\mathrm{T}=2 \mathrm{sec}$

So, $A=\frac {2mg}{K}$

${\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}}  {\text { or } \quad \mathrm{k}_{2}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$

${\mathrm{Now} \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_{2}}}}  {\text { or } \quad \mathrm{k}=\frac{4 \pi^{2} \mathrm{m}}{\mathrm{T}_{2}^{2}}}$

In parallel $\mathrm{k}=\mathrm{k}_{1}+\mathrm{k}_{2}$

Substituting the values of $\mathrm{k}, \mathrm{k}_{1}$ and $\mathrm{k}_{2}$ we get:

$\frac{1}{\mathrm{T}^{2}}=\frac{1}{\mathrm{T}_{1}^{2}}+\frac{1}{\mathrm{T}_{2}^{2}}$

$\mathrm{k}_{\mathrm{eq}}=\left[\frac{1}{2 \mathrm{k}_{1}}+\frac{1}{\mathrm{k}_{2}}\right]^{-1}$

$\therefore \frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\sqrt{\frac{\mathrm{m}_{2}}{\mathrm{m}_{1}}}=\sqrt{\frac{400}{900}}$

$\Rightarrow \frac{\mathrm{T}_{2}}{3}=\frac{2}{3} \Rightarrow \mathrm{T}_{2}=2 \mathrm{s}$

$\omega=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}=\sqrt{\frac{1200}{3}}=20 \mathrm{rad} / \mathrm{sec}$

or $10=20 \sin \omega t$

$\therefore \omega t=\left(\frac{\pi}{6}\right) \Rightarrow t=\frac{\pi}{6 \omega}$

The desired time will be $2 t$

or $\frac{\pi}{3 \omega}=\frac{\pi}{3(2 \pi / T)}=\frac{T}{6}=\frac{12}{6}=2 s$

$\frac{1}{K_{e q}}=\frac{1}{K}\left[\frac{1}{1-\frac{1}{2}}\right]=\frac{2}{K} \Rightarrow K_{e q}=\frac{K}{2}$

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