Oscillations Practice Questions

At any instant the total energy is$\;\frac{1}{2} k A^2=$ constant, where $A=$ amplitudehence total energy is independent of $x$.

$t=2 \pi \sqrt{\;\frac{ℓ}{g_{e f f}}} ; t_o=2 \pi \sqrt{\;\frac{ℓ}{g}}$ $m g_{\;\text{eff }\;}=m g-B=m y-V \times 100 \times g$ $\therefore g_{\;\text{eff }\;}=g-\;\frac{100}{(m / v)} g$ $=g-\;\frac{1000}{\frac{4}{3} \times 1000} g=\;\frac{g}{4}$ $\therefore t=2 \pi \sqrt{\;\frac{ℓ}{g {/} 4}} \;\; t=2 t_0$

The maximum of amplitude and energy is obtained when the frequency is equal to the natural frequency (resonance condition) $\therefore \omega _1=\omega _2$

Given that, initial angular velocity $=\omega _0$ and at any instant time $t$, angular velocity $=\omega$ So when displacement is ${\text{x}}$ then the resultant acceleration ${\text{f}}= (\omega _0^{2-}\omega ^2) x$ So the external force, ${\text{F}}=m (\omega _0^{2-}\omega ^2) x$....(1) But given that $F \propto \cos \omega t$ From (i) we get, $m (\omega _0^{2-}\omega ^2) x \propto \cos \omega t$....(ii) From equation of SHM we know, $x=A sin \;(\omega t+\varphi )$ When ${\text{t}}=0$ then ${\text{x}}= {\text{A}}$ $\therefore {\text{A}}=A sin \;(\varphi )$ $\Rightarrow A=\;\frac{\pi }{2}$ $\therefore x=A sin \; (\omega t+\;\frac{\pi }{2}) =A \cos \omega t$Putting value of $x$ in (ii), we get$\;m (\omega _0^{2-}\omega ^2) A \cos \omega t \propto \cos \omega t$ $\;\Rightarrow A \propto \;\frac{1}{m (\omega _0^{2-}\omega ^2) }$

$K . E .=\;\frac{1}{2} m \omega ^2 (a^{2-}x^2)$When $x=0, K$. $E$ is maximum and is equal to $\;\frac{1}{2} m \omega ^2 a^2$.

$x=4(\cos \pi t+ sin \;\pi t)$ $={\sqrt{2}} \times 4 (\;{ sin \;\pi t}/{{\sqrt{2}}}+\;{\cos \pi t}/{{\sqrt{2}}})$ $x=4 {\sqrt{2}} sin \; (\pi t+45^{\circ})$

$T=2 \pi \sqrt{\;\frac{ℓ}{g}}$ and $T^{'}=2 \pi \sqrt{\;\frac{1.21 ℓ}{g}}$ $(.$ as $ℓ^{'}=ℓ+21 \%$ of $ℓ)$ $\%$ increase $=\;\frac{T^{'}-T}{T} \times 100$ $=\;{{\sqrt{1.21 ℓ}}-{\sqrt{ℓ}}}/{{\sqrt{ℓ}}} \times 100$ $=\;{\sqrt{\frac{121}{10} l} l-{\sqrt{l}}}/{{\sqrt{l}}} \times 100$ $=\;{\;\frac{11}{10} {\sqrt{l}}-{\sqrt{l}}}/{{\sqrt{l}}} \times 100$ $=\;\frac{\;\frac{11}{10}-1}{1} \times 10$ $=10 \%$

$\;T=2 \pi \sqrt{\;\frac{M}{k}}$......(i)$\;T^{'}=2 \pi \sqrt{\;\frac{M+m}{k}}$ $\;\Rightarrow \;\frac{5 T}{3}=2 \pi \sqrt{\;\frac{M+m}{k}} .$....(ii)Dividing Eq. (i) by Eq. (ii), we have$\; \therefore \;\frac{3}{5}=\sqrt{\;\frac{M}{M+m}}$ $\;\;\frac{9}{25}=\;\frac{M}{M+m}$ $\;\Rightarrow 9 M+9 m=25 M$ $\;\Rightarrow 16 M=9 m$ $\;\;\frac{m}{M}=\;\frac{16}{9}$

Maximum velocity during $S H M=A \omega =A \sqrt{\;\frac{k}{m}}$ $[ \therefore \omega =\sqrt{\;\frac{k}{m}}]$Here the maximum velocity is same and $m$ is also same$\; \therefore A_1 {\sqrt{k_1}}=A_2 {\sqrt{k_2}}$ $\; \therefore \;\frac{A_1}{A_2}=\sqrt{\;\frac{k_2}{k_1}}$

Let the spring constant of the original spring be $k$. Then its time period $T=2 \pi \sqrt{\;\frac{m}{k}}$ where $m$ is the mass of oscillating body. When the spring is cut into $n$ equal parts, the spring constant of one part becomes $n k$. Therefore the new time period, $T^{'}=2 \pi \sqrt{\;\frac{m}{n k}}=\;\frac{T}{\sqrt{n}}$

KEY CONCEPT : The time period $T=2 \pi \sqrt{\;\frac{ℓ}{g}}$ where $ℓ=$ distance between the point of suspension and the center of mass of the child. This distance decreases when the child stands $\therefore T^{'}< T$ i.e., the period decreases.

$K . E=\;\frac{1}{2} k(A^{2-}x^2) ; \;\; U=\;\frac{1}{2} k x^2$ At the mean position $x=0$ $\therefore K . E .=\;\frac{1}{2} k A^2=$ Maximum and $U=0$