Oscillations Practice Questions

$\;\frac{1}{2} m v^2=m g l(1-\cos \theta )$ $\Rightarrow v={\sqrt{2 g l(1-\cos \theta )}}$ $=\sqrt{2 \times 10 \times 2.5 \times \;\frac{1}{2}}$ $=5 {\text{m}} / {\ s}$

$\;\frac{1}{2} {\text{kA}}^2=\;{ {\text{p}}^2}/{2 {\text{m}}}$ $\Rightarrow (\;{ {\text{A}}_1}/{ {\text{A}}_2})^2=\;{ {\text{m}}_2}/{ {\text{m}}_1}=\;\frac{1024}{900}$ $\Rightarrow \;{ {\text{A}}_1}/{ {\text{A}}_2}=\;\frac{32}{30}=\;\frac{16}{15}=\;\frac{16}{16-1}$ $\therefore \alpha =16$

$U=4(1-\cos 4 x)$ $F=-\;\frac{d U}{d x}=-4(+\ sin 4 x) 4=-16 \ sin (4 x)$For small $\theta$ $\ sin \theta \approx \theta$ ${\text{F}}=-64 {\text{x}}$ $a=-64 x / m=-16 x$ $\omega ^2=16$ ${\text{T}}=\;\frac{2 \pi }{\omega }=\;\frac{\pi }{2}$

$\omega =\pi =\sqrt{\;\frac{g}{l}}$So, $l=\;\frac{g}{\pi ^2}$ $≃ 99.4 {\text{cm}}$ (Nearest value)

$v=\omega {\sqrt{A^{2-}y^2}}$ $\Rightarrow 3 \omega {\sqrt{10^{2-}5^2}}=\omega \sqrt{ (A^{'}) ^{2-}5^2}$ $\Rightarrow 9 \times 75= (A^{'}) ^{2-}25$ $\Rightarrow A^{'}={\sqrt{28 \times 25}} {\text{cm}}$ $\Rightarrow x=700$

Let $x=A \sin \omega t$ $\Rightarrow v=A \omega \cos \omega t$ $\Rightarrow v=± \omega \sqrt{A^{2-}x^2}$ $\Rightarrow \;\frac{v^2}{\omega ^2}+x^2=A^2$ $\Rightarrow \;\text{ Ellipse }\;$

Both the springs are in parallel combination in both the diagrams so$T_{1}=2 \pi \sqrt{\;\frac{3 m}{2 k}}$ $\;\text{ and }\; T_{2}=2 \pi \sqrt{\;\frac{m}{3 k}}$So, $\;\frac{T_{1}}{T_{2}}=\;\frac{3}{\sqrt{2}}$

$\omega _1 A_1=\omega _2 A_2$ $\Rightarrow \;\frac{A_1}{A_2}=\;\frac{\omega _2}{\omega _1}$ $=\sqrt{\;\frac{k_2}{m_2}} \times \sqrt{\;\frac{m_1}{k_1}}=\sqrt{\;\frac{9 k}{100} \times \;\frac{50}{2 k}}=\;\frac{3}{2}$

${\text{g}}_{ {\text{cff}}}= {\text{g}} \cos \alpha$ ${\text{T}}=2 \pi \sqrt{\;{ {\text{L}}}/{ {\text{g}} \cos \alpha }}$

$x= sin \; (\pi t+\;\frac{\pi }{3}) m$ $\Rightarrow \;\frac{d x}{d t}=\pi \cos (\pi t+\;\frac{\pi }{3})$ $=\pi \cos (\pi +\;\frac{\pi }{3}) \;\text{ at }\; t=1 s$ $=-\;\frac{\pi }{2} {\text{m}} / {\ s}$or $|\;\frac{d x}{d t}{{}/{}|} =157 {\text{cm}} / {\ s}$

For figure (a) :${\text{K}}_{\;\text{eq }\;}=\;{ {\text{K}} \times 2 {\text{K}}}/{ {\text{K}}+2 {\text{K}}}=\;{2 {\text{K}}}/{3}$ ${\text{T}}=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{K}}_{ {\text{eq}}}}}=2 \pi \sqrt{\;{ {\text{m}}}/{2 {\text{K}} / 3}}=2 \pi \sqrt{\;{3 {\text{m}}}/{2 {\text{K}}}}$For figure (b):${\text{K}}_{ {\text{eq}}}=3 {\text{K}}, {\text{T}}^{'}=2 \pi \sqrt{\;{ {\text{m}}}/{3 {\text{K}}}}$ $\;{ {\text{T}}^{'}}/{ {\text{T}}}=\sqrt{\;{ {\text{m}} \times 2 {\text{K}}}/{3 {\text{K}} \times 3 {\text{m}}}}=\;\frac{\sqrt{2}}{3}$ ${\text{T}}^{'}=\sqrt{2}$ ${\text{x}}=2$

$\omega =\\sqrt{{{k}_{{eq}}}/{\mu}}$ $\mu=$ reduced mass springs are in series connection ${k}_{{eq}}=\frac{{k}_{1} {k}_{2}}{{k}_{1}+{k}_{2}}$ ${k}_{{eq}}=\frac{{k} \\times 4 {k}}{5 {k}}=\frac{4 {k}}{5}$ ${k}_{{eq}}=\frac{4 \\times 20}{5}{N} / {m}=16 {N} / {m}$ $\mu=\frac{{m}_{1} {m}_{2}}{{m}_{1}+{m}_{2}}=\frac{0.2 \\times 0.8}{0.2+0.8}$ $=0.16 {kg}$ $\omega =\\sqrt{\frac{16}{0.16}}=\\sqrt{100}=10$

Given that, a bob of mass m suspended by a thread of length l undergoes SHM with time period, $\text{T}_{1}=\text{T}=2 \pi \\sqrt{\frac{l}{g}}$ ...(i) In liquid, effective gravity, $g_{{\text{eff} }}=g(\frac{\sigma -\rho }{\sigma })$ where, ρ = density of liquid and σ = density of body (bob). Given, $\rho =\frac{\sigma }{4}$ $\Rightarrow \sigma =4 \rho$ Substituting this value, we get $g_{{\text{eff}}}=g(\frac{4 \rho -\rho }{4 \rho })=\frac{3 g}{4}$ Now, the new time period of pendulum in liquid is $\text{T}'=2 \pi \\sqrt{{\text{l}'}/{g_{{\text{eff}}}}}$ ...(ii) Here,$l'=l+\frac{l}{3}=\frac{4 \text{I}}{3}$ Substituting the values in Eq. (ii), we get $\text{T}'=2 \pi \\sqrt{\frac{4 l}{3} \times \frac{4}{3 g}}$ $=(2 \pi \\sqrt{\frac{l}{g}}) \frac{4}{3}$ $\Rightarrow \text{T}'=\frac{4 \text{T}}{3}$ [from Eq. (i)]

Given, length of simple pendulum, $I=2 {m}$ Time period, $T=2 {s}$ Let $g_{\;\text{eff }\;}$ be the acceleration due to gravity. $\therefore \;\text{ Time period, }\; T \;\; =2 \pi \sqrt{\;{1}/{g_{\;\text{eff }\;}}}$ $\Rightarrow \;\; g_{\;\text{eff }\;} \;\; =4 \pi ^{2} \;\frac{1}{T^{2}}$ $\;\; =4 \pi ^{2} \cdot \;\frac{2}{4}=2 \pi ^{2} {ms}^{-2}$

Given, $T=2 \pi \sqrt{\;\frac{L}{g}}$ $. . .$ (i) where, time period, $T=1.95 {s}$ Length of string, $I=1 {m}$ Acceleration due to gravity $=g$ Error in time period, $\Delta T=0.01 {s}=10^{-2} {s}$ Error in length, $\Delta L=1 {mm}=1 \times 10^{-3} {m}$ Squaring Eq. (i) on both sides, we get $T^{2} \;\; =4 \pi ^{2} \;\frac{L}{g}$ $g \;\; =4 \pi ^{2} \;\frac{L}{T^{2}}$ $\Rightarrow \;\; \;\frac{\Delta g}{g} \;\; =\;\frac{\Delta L}{L}+\;\frac{2 \Delta T}{T}=\;\frac{10^{-3}}{1}+\;\frac{2 \times 10^{-2}}{1.95}$ $\;\; =10^{-3}+1.025 \times 10^{-2}$ $\;\; =10^{-3}+10.25 \times 10^{-3}$ $\;\; =11.25 \times 10^{-3}$ $\because \;\; \Delta g / g \times 100 \;\; =11.25 \times 10^{-3} \times 10^{2}$ $\;\; =1.125 \%=1.13 \%$

Velocity of particle in ${SHM}$ in terms of displacement $x$ is given as $v \;\; =\omega \sqrt{A^{2}-x^{2}}$ $\Rightarrow$ $v \;\; =A \omega \sqrt{1-x^{2} / A^{2}}$ $\Rightarrow$ $\;\frac{v}{A \omega } \;\; =\sqrt{1-x^{2} / A^{2}}$ $\Rightarrow$ $\;\frac{v^{2}}{A^{2} \omega ^{2}} \;\; =1-\;\frac{x^{2}}{A^{2}}$ $\Rightarrow$ $\;\frac{v^{2}}{A^{2} \omega ^{2}}+$ $\;\frac{x^{2}}{A^{2}} \;\; =1$ . . . (i) As equation of ellipse is $\;\frac{x^{2}}{a^{2}}+\;\frac{y^{2}}{b^{2}}=1$ Eq. (i) is similar to the equation of ellipse. So, the graph between velocity and displacement is an ellipse.

From the potential energy curve, $\text{U}_{\max }=\frac{1}{2} \text{k} \text{A}^{2}$ $10=\frac{1}{2} \text{k}(2)^{2}$ \Rightarrow k = 5 N / m The length of the simple pendulum, L = 4 m Time period of spring, $\text{T}=2 \pi \\sqrt{\frac{k}{m}}$ Time period of simple pendulum, $\text{T}=2 \pi \\sqrt{\frac{\text{I}}{g}}$ The time period of simple pendulum is same as the time period ofthe spring oscillation. $\Rightarrow 2 \pi \\sqrt{\frac{\text{I}}{g}}=2 \pi \\sqrt{\frac{k}{m}}$ Substituting the values in the above equations, we get $2 \pi \\sqrt{\frac{4}{g}}=2 \pi \\sqrt{\frac{5}{5}}$ $g=4 {\text{m}} / {\text{s}}^{2}$ ∴ The acceleration due to the gravity on the planet is $4 {\text{m}}/ {\text{s}}^{2}$.

Let series equivalent of spring constant $=k_{ {eq}}$ and $T$ be the time period. In series arrangement, $\;\frac{1}{k_{e q}}=\;\frac{1}{k_{1}}+\;\frac{1}{k_{2}}$ $\Rightarrow \;\; \;{1}/{k_{ {eq}}}=\;\frac{1}{k_{1}}+\;\frac{1}{k_{1}}=\;\frac{2}{k_{1}} \Rightarrow k_{ {eq}}=\;\frac{k_{1}}{2}$ As, $\;\; T=2 \pi \sqrt{\;\frac{m}{k_{1}}}$ where, $m$ is mass of body connected with spring. $\Rightarrow \;\; T \propto \sqrt{\;\frac{1}{k_{1}}}$ and $\;\; T^{'} \propto \sqrt{\;\frac{2}{k_{1}}} \Rightarrow T^{'}=\sqrt{2} T$