Physics 3 Marks Question

$\therefore$ Resultant motion will be given by,

$\text{R}=\sqrt{(\text{x}^2+\text{s}^2+2\text{xs}\cos45^\circ)}$

$=\sqrt{\Big\{\text{x}_0^2\sin^2\omega\text{t}+\text{s}_0^2\sin^2\omega\text{t}+2\text{x}_0\text{s}_0\sin^2\omega\text{tx}\Big(\frac{1}{\sqrt{2}}\Big)\Big\}}$

$=\big[\text{x}_0^2+\text{s}_0^2=\sqrt{2}\text{x}_0\text{s}_0\big]^{\frac{1}{2}}\sin\omega\text{t}$

$\therefore$ Resultant amplitude $=\big[\text{x}_0^2+\text{s}_0^2=\sqrt{2}\text{x}_0\text{s}_0\big]^{\frac{1}{2}}$

$\therefore$ Total Resultant force $=\text{kx}+\sqrt{\Big(\frac{\text{kx}}{\sqrt{2}}\Big)^2+\Big(\frac{\text{kx}}{\sqrt{2}}\Big)^2}=\text{kx}+\text{kx}=2\text{kx}.$

$\Rightarrow\text{x}=\frac{2\ell}{3}$

$\therefore$ Length of the simple pendulum $=\frac{2\ell}{3}$

$\Rightarrow\text{mv}^2+2\text{k}\text{x}^2=\text{c}.$

$\text{m}\times2\text{v}\frac{\text{dv}}{\text{dt}}+2\text{k}\times2\text{x}\frac{\text{dx}}{\text{dt}}=0$

$\therefore\text{ma}+2\text{kx}=0$ $\Big[$ because $\text{v}=\frac{\text{dx}}{\text{dt}}$ and $\text{a}=\frac{\text{dv}}{\text{dt}}\Big]$

$\Rightarrow\frac{\text{a}}{\text{x}}=-\frac{2\text{k}}{\text{m}}=\omega^2$

$\Rightarrow\omega=\sqrt{\frac{2\text{k}}{\text{m}}}$

$\Rightarrow$ Time period $\text{T}=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}$

$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{k}}}=2\pi\sqrt{\frac{\text{mr}^2}{2\text{k}}}$ [where K = Torsional constant]

$\text{T}^2=4\pi^2\frac{\text{mr}^2}{2\text{K}}=2\pi^2\frac{\text{mr}^2}{\text{k}}$

$\Rightarrow2\pi^2\text{mr}^2=\text{KT}^2$

$\Rightarrow\text{K}=\frac{2\text{mr}^2\pi^2}{\text{T}^2}$

$\therefore$ Torsional constant $\text{K}=\frac{2\text{mr}^2\pi^2}{\text{T}^2}$

$\therefore\text{a}=\frac{\text{F}}{\text{m}}=\frac{3\text{kx}}{2\text{m}}$

$\Rightarrow\frac{\text{a}}{\text{x}}=\frac{3\text{k}}{2\text{m}}=\omega^2$

$\Rightarrow\omega=\sqrt{\frac{3\text{k}}{2\text{m}}}$

$\therefore$ time period $\text{T}=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{2\text{m}}{3\text{k}}}$

$\therefore$ Time taken by the block to move from M → N and N → M

$\text{v}^2=2\text{g}\ell(1-\cos\theta)$

$\text{E}=5\text{J}$

$\text{f}=5$

So, $\text{T}=\frac{1}{5}\sec.$

Now, $\text{P.E}=\Big(\frac{1}{2}\Big)\text{kx}^2$

$\Rightarrow\Big(\frac{1}{2}\Big)\text{kx}^2=5$

$\Rightarrow\Big(\frac{1}{2}\Big)\text{k}(0.25)^2=5$

$\Rightarrow\text{k}=160\text{N/m}.$

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

$\Rightarrow\frac{1}{5}=2\pi\sqrt{\frac{\text{m}}{160}}$

$\Rightarrow\text{m}=0.16\text{kg.}$

$\text{T}=4\sec.$

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

$\Rightarrow4=2\pi\sqrt{\frac{2}{\text{k}}}$

$\Rightarrow2=\pi\sqrt{\frac{2}{\text{k}}}$

$\Rightarrow4=\pi^2\Big(\frac{\text{2}}{\text{k}}\Big)$

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

$\Rightarrow\text{k}=\frac{\pi^2}{2}$

$\Rightarrow\text{k}=\frac{\pi^2}{2}=5\text{N/m}$

But, we know that $\text{F}=\text{mg}=\text{kx}$

$\Rightarrow\text{x}=\frac{\text{mg}}{\text{k}}=\frac{2\times10}{5}=4$

$\therefore$ Potential Energy $=\big(\frac{1}{2}\big)\text{kx}^2=\big(\frac{1}{2}\big)\times5\times16=5\times8=40\text{J}$

$\text{T}=0.314\sec$

$\text{m}=0.5\text{kg}.$

Total force exerted on the block = weight of the block + spring force.

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

$\Rightarrow0.314=2\pi\sqrt{\frac{0.5}{\text{k}}}$

$\text{k}=200\text{N/m}$

$\therefore$ Force exerted by the spring on the block is $\text{F}=\text{kx}=201.1\times0.1=20\text{N}$

$\therefore$ Maximum force $=\text{F}+\text{Weight}=20+5=25\text{N}$

$\Rightarrow\text{r}\omega=10$

$\Rightarrow\omega^2=\frac{100}{\text{r}^2}\ ...(1)$

$\text{A}_{\text{max}}=\omega^2\text{r}=50\text{cm/sec}$

$\Rightarrow\omega^2=\frac{50}{\text{y}}=\frac{50}{\text{r}}\ ...(2)$

$\therefore\frac{100}{\text{r}^2}=\frac{50}{\text{r}}$

$\Rightarrow\text{r}=2\text{cm}.$

$\therefore\omega=\sqrt{\frac{100}{\text{r}^2}}=5\sec^2$

Again, to find out the positions where the speed is 8m/sec,

$\text{v}^2=\omega^2(\text{r}^2-\text{y}^2)$

$\Rightarrow64=25(4-\text{y}^2)$

$\Rightarrow4-\text{y}^2=\frac{64}{25}$

$\Rightarrow\text{y}^2=1.44$

$\Rightarrow\text{y}=\sqrt{1.44}$

$\Rightarrow\text{y}=\pm1.2\text{cm}$ from mean position.

Now, $\text{y}_1=\text{r}\sin\omega\text{t}_1$

$\Rightarrow\frac{\text{r}}{2}=\text{r}\sin\omega\text{t}_1$

$\Rightarrow\sin\omega\text{t}_1=\frac{1}{2}$

$\Rightarrow\omega\text{t}_1=\frac{\pi}{2}$

$\Rightarrow\frac{2\pi}{\text{t}}\times\text{t}_1=\frac{\pi}{6}$

$\Rightarrow\text{t}_1=\frac{\text{t}}{12}$

Again, $\text{y}_2=\text{r}\sin\omega\text{t}_2$

$\Rightarrow\text{r = r}\sin\omega\text{t}_2$

$\Rightarrow\sin\omega\text{t}_2=1$

$\Rightarrow\omega\text{t}_2=\frac{\pi}{2}$

$\Rightarrow\Big(\frac{2\pi}{\text{t}}\Big)\text{t}_2=\frac{\pi}{2}$

$\Rightarrow\text{t}_2=\frac{\text{t}}{4}$

So, $\text{t}_2-\text{t}_1=\frac{{\text{t}}{}}{4}-\frac{\text{t}}{12}=\frac{\text{t}}{6}$