Oscillations Practice Questions

$\; {\text{F}}=-2 {\text{kx}}, {\text{a}}=-\;{2 {\text{kx}}}/{ {\text{m}}}, \omega =\sqrt{\;{2 {\text{k}}}/{ {\text{m}}}}=\sqrt{\;\frac{2 \times 20}{2}}$ $\;={\sqrt{20}} {\text{rad}} / {\ s}$ $\; {\text{T}}=\;\frac{2 \pi }{\omega }=\;{2 \pi }/{{\sqrt{20}}}=\;{\pi }/{{\sqrt{5}}}$ $\; {\text{x}}=5$

$\;K E=P E+\;\frac{P E}{4}$ $\;K E=\;\frac{5}{4} P E$ $\;\;\frac{1}{2} m \omega ^2 (A^{2-}x^2) =\;\frac{5}{4} \times \;\frac{1}{2} m \omega ^2 x^2$ \;{[ {\text{v}}=\omega \sqrt{A^{2-} {\text{x}}^2}]\ }$$;A^{2-}x^2=;\frac{5}{4} x^2$$;;\frac{9 x^2}{4}=A^2$$;x=;\frac{2}{3} A$$; ∴ x=;\frac{2}{3} × 3 {\text{cm}}$$;x=2 {\text{cm}}$

At surface of earth time period${\text{T}}=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$At height ${\text{h}}= {\text{R}}$ $\; {\text{g}}^{'}=\;{ {\text{g}}}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}=\;{ {\text{g}}}/{4}$ $\; {\text{xT}}=2 \pi \sqrt{\;{ℓ}/{( {\text{g}} / 4)}}$ $\;\Rightarrow {\text{xT}}=2 \times 2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\;\Rightarrow {\text{xT}}=2 {\text{T}} \Rightarrow {\text{x}}=2$

Let time from 0 to ${\text{A}} / 2$ is ${\text{t}}_1$ $\&$ from ${\text{A}} / 2$ to ${\text{A}}$ is ${\text{t}}_2$then $\omega {\text{t}}_1=\pi / 6$ $\omega t_2=\pi / 3$ $\;\frac{t_1}{t_2}=\;\frac{1}{2}$ $t_2=2 t_1=2 \times 2=4 {\ sec}$

In SHM,$\; {\text{F}} \propto - {\text{x}} \to {\text{A}}$ $\; {\text{a}} \propto - {\text{x}} \to {\text{B}}$ $\; {\text{V}}_{\;\text{mean }\;} \to \;\text{ max imum }\; \to {\text{c}}$ $\; {\text{a}}_{\;\text{extreme }\;} \to \;\text{ maximum }\;$Hence, (A), (B) and (C) are true.

$x=A sin \;(\omega t)$ ${\text{U}}_{( {\text{x}})}=\;\frac{1}{2} {\text{kx}}^2$ $\;{ {\text{dU}}}/{ {\text{dt}}}=\;\frac{1}{2} {\text{k}} 2 {\text{x}} \;{ {\text{dx}}}/{ {\text{dt}}}$ $= {\text{kA}}^2 \omega sin \;\omega {\text{t}} \cos \omega {\text{t}} \times \;\frac{2}{2}$ $\; (\;{ {\text{dU}}}/{ {\text{dt}}}) _{\max }=\;{ {\text{kA}}^2 \omega }/{2}( sin \;2 \omega {\text{t}})_{\max }$ $\;2 \omega {\text{t}}=\;\frac{\pi }{2} \Rightarrow {\text{t}}=\;\frac{\pi }{4} \omega =\;{ {\text{T}}}/{8} \Rightarrow \beta =8$

Both the springs are in parallel.$\;K_{e q}=K_1+K_2$ $\;T=2 \pi \sqrt{\;\frac{m}{K_{e q}}}=2 \pi \sqrt{\;\frac{m}{K_1+K_2}}$

$\; {\text{T}}=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{k}}}}=1$ $\; {\text{T}}^{'}=2 \pi \sqrt{\;{ {\text{m}}+3}/{ {\text{k}}}}=2$ $\;\;{ {\text{T}}}/{ {\text{T}}^{'}}=\sqrt{\;{ {\text{m}}}/{ {\text{m}}+3}}=\;\frac{1}{2}$ $\;\Rightarrow \;{ {\text{m}}}/{ {\text{m}}+3}=\;\frac{1}{4}$ $\; {\text{m}}=1$

Ratio of ${\text{PE}}$ and ${\text{KE}}=$ ?When $\;{ {\text{x}}}_{{{ {\text{x}}} \\ \;\text{ disp. }\;}}=\;{ {\text{A}}}/{2}$ ${\text{KE}}\;=\;\frac{1}{2} {\text{m}} \omega ^2 [ {\text{A}}^{2-} {\text{x}}^2]$ $\;=\;\frac{1}{2} {\text{m}} \omega ^2 [ {\text{A}}^{2-}\;{ {\text{A}}^2}/{4}]$ $\; {\text{KE}}==\;\frac{1}{2} {\text{m}} \omega ^2 [\;{3 {\text{A}}^2}/{4}]$ ... (1)$\;\;\text{ And }\; {\text{PE}}=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{x}}^2$ $\;=\;\frac{1}{2} {\text{m}} \omega ^2 \;{ {\text{A}}^2}/{4}$ $\;\;{ {\text{PE}}}/{ {\text{KE}}}=\;{ {\text{m}} \omega ^2 {\text{A}}^2 \times 8}/{8 {\text{m}} \omega ^2 {\text{A}}^2 \times 3}=\;\frac{1}{3}$ $\;\Rightarrow 1: 3$ $\;$

$\; {\text{u}}_{\max }=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2=25 {\text{J}}$ $\; {\text{KE}} \;\text{ at }\; \;{ {\text{A}}}/{2}=\;\frac{1}{2} {\text{m}} {\text{v}}_1^2=\;\frac{1}{2} {\text{m}} \omega ^2 ( {\text{A}}^{2-}\;{ {\text{A}}^2}/{4})$ $\; {\text{KE}}=\;\frac{1}{2} {\text{m}} \omega ^2 \;{3 {\text{A}}^2}/{4}=\;\frac{3}{4} (\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2)$ $\; {\text{KE}}=\;\frac{3}{4} \times 25=18.75 {\text{J}}$

$\;\omega =\sqrt{\;{ {\text{k}}}/{ {\text{m}}}}$ $\;\;\frac{\omega _2}{\omega _1}=\sqrt{\;{ {\text{m}}_1}/{ {\text{m}}_2}}=\sqrt{\;\frac{1}{2}}$

$\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-} {\text{x}}^2}$ $\; {\text{y}}= {\text{A}} \times \;{{\sqrt{3}}}/{2}$ $\; {\text{v}}=\omega \sqrt{ {\text{A}}^{2-}\;{3 {\text{A}}^2}/{4}}=\;{\omega {\text{A}}}/{2}$ $\;\;\text{ Given, }\; {\text{KE}}=200 {\text{J}}$ $\;\;\frac{1}{2} {\text{m}} \;{\omega ^2 {\text{A}}^2}/{4}=200$ $\; {\text{KA}}{ }^2=1600 \;\; ( {\text{K}}= {\text{m}} \omega ^2)$ $\; {\text{K}}=\;\frac{1600}{ (40 \times 10^{-2}) ^2}$ $\; {\text{K}}=10^4 {\text{N}} / {\text{m}}$ $\; {\text{x}}=4$

$\; {\text{K}} . {\text{E}}= {\text{P}} . {\text{E}}$ $\;\;\frac{1}{2} M \omega ^2 (A^{2-}x^2) =\;\frac{1}{2} M \omega ^2 x^2$ $\; (A^{2-}x^2) =x^2$ $\;2 x^2=A^2$ $\;x= ± \;{A}/{{\sqrt{2}}}$

When an object executes S.H.M, its morning acceleration is given by ${\text{a}}_{\max }=\omega ^2 {\text{A}}$Where $\omega =\;\frac{2 \pi }{T}$Therefore, ${\text{a}}_{\max }=\;{4 \pi ^2 {\text{A}}}/{ {\text{T}}^2}$(Max force) ${\text{F}}_{\max }= {\text{ma}}_{\max }=5 \times \;\frac{4 \times 3.14 \times 3.14}{3.14 \times 3.14} \times (1)$ $=20 {\text{N}}$

$x=4 sin \; (\;\frac{\pi }{2}-\omega t)$ $=4 \cos (\omega t)$ $y=4 sin \;(\omega t)$ $\Rightarrow x^{2+}y^2=4^2$ $\Rightarrow$ The particle is moving in a circular motion with radius of $4 {\text{m}}$.

${\text{A}}=8 {\text{cm}}$ ${\text{T}}=6 {\ s}$ $A \cos (\;\frac{2 \pi t}{T}) =\;\frac{A}{2}$ $\Rightarrow \;\frac{2 \pi t}{T}=\;\frac{\pi }{3}$or $t=\;\frac{T}{6}=1 s$

$T \propto \sqrt{1 / g}$ $\Rightarrow \;\frac{T_1}{T_2}=\sqrt{\;\frac{g_2}{g_1}}=\;\frac{R}{R+h}$ $\;\frac{4}{6}=\;\frac{R}{R+h}$ $\Rightarrow h=R / 2$ $=3200 {\text{km}}$

${\text{mg}}^{'}= {\text{mg}}- {\text{F}}_{ {\text{B}}}$ ${\text{g}}^{'}=\;{ {\text{mg}}- {\text{F}}_{ {\text{B}}}}/{ {\text{F}}_{ {\text{B}}}}$ $=\;{\rho _{ {\text{B}}} {\text{Vg}}-\rho _{ {\text{w}}} {\text{Vg}}}/{\rho _{ {\text{B}}} {\text{V}}}$ $= (\;{\rho _{ {\text{B}}}-\rho _{ {\text{w}}}}/{\rho _{ {\text{B}}}}) {\text{g}}$ $=\;\frac{5-1}{5} \times {\text{g}}$ $=\;\frac{4}{5} {\text{g}}$We know, $T=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\;{ {\text{T}}^{'}}/{ {\text{T}}}=\sqrt{\;{ {\text{g}}}/{ {\text{g}}^{'}}}=\sqrt{\;{ {\text{g}}}/{5} {\text{g}}}=\sqrt{\;\frac{5}{4}}$ ${\text{T}}^{'}= {\text{T}} \sqrt{\;\frac{5}{4}}=\;\frac{10}{2} {\sqrt{5}}$ ${\text{T}}^{'}=5 {\sqrt{5}}$