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}$ 

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

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

$ = 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 $S H M: x=A \sin (\omega t+\phi)$

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

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

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

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

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

$ \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 $=a c o t$

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

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

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