Oscillations Practice Questions

Given, mass block, $m=1 {kg}$ Initial amplitude, $A_{0}=12 {cm}$ Final amplitude, $A=6 {cm}$ The time taken to reduce the amplitude, $t=2 {min}=120 {s}$ Using the expression of damped oscillation, $A=A_{0} e^{-\;\frac{b}{2 m} t}$ Substituting the values in the above equation, we get $6=12 e^{-\;\frac{b(120)}{2(1)}}$ $\Rightarrow \;\; e^{60 b}=2$ or $60 b=10 \log 2$ $\Rightarrow \;\; b=\;\frac{0.693}{60}=1.16 \times 10^{-2} \frac{kg}{s}$

Given, initial amplitude $=A$ Velocity at mean position, $v=A \omega$ Applying conservation of momentum at mean position, we get $M_{1} v_{1} \;\; =M_{2} v_{2}$ $M A \omega \;\; =(M+m) v^{'}$ $\Rightarrow \;\; v^{'} \;\; =\;\frac{M A \omega }{M+m}=\;{M A \sqrt{\frac{k}{M}}}/{M+m}$ $\;\; v^{'} \;\; =A^{'} \omega ^{'}=A^{'} \sqrt{\;\frac{k}{M+m}}$ $\therefore \;\; A^{'} \;\; =\;{M A \sqrt{\frac{k}{M}}}/{M+m} \times \sqrt{\;\frac{M+m}{k}}$ $A^{'} \;\; =\sqrt{\;\frac{M}{M+m}} A$

Given, Mass, $m=500 {g}$ Decay constant, $\lambda =20 {g} / {s}$ For damped oscillation, $A=A_{0} {e}^{-b {bt} / 2 m}$....(i) Here, $b=$ damped constant $=$ decay constant As per question, the amplitude of the system is dropped to half of its initial value. So, Eq. (i) becomes $\;\frac{A_{0}}{2}=A_{0} {e}^{-b t / 2 m} \Rightarrow \;\frac{1}{2}= {e}^{-b t / 2 m}$ $\Rightarrow \;\; 2= {e}^{b t / 2 m}$ $\Rightarrow \;\; 2= {e}^{b t / 2 m}$ Taking log on both sides of above equation, we get $\log 2=\log (e^{b t / 2 m})$ $\Rightarrow \;\; \log 2=\;\frac{b t}{2 m} \Rightarrow t=\;\frac{2 m}{b} \log 2$ $=\;\frac{2 \times 500 \times 0.693}{20}$ $[ \because \log 2=0.693]$ $\Rightarrow t=34.65 {s}$

Applying work energy theorem: ${W}_{{g}}+{w}_{{T}}=\Delta{K}$ $-{mgl}(1-\cos 60°)=\frac{1}{2} {mv}^{2}-\frac{1}{2} {\mu}^{2}$ $v^{2}=u^{2}-2 g l(1-\cos 60°)$ ${v}^{2}=9-2 \\times 10 \\times 0.5(\frac{1}{2})$ ${v}^{2}=4$ ${v}=2 {m} / {s}$

Given, the measured value of the length of pendulum, $I=10 {cm}$ The time taken to complete the 200 oscillations, $T=100 {s}$ Time resolution, $\Delta T=1 {s}$ Accuracy in length, $\Delta l=1 {mm}=0.1 {cm}$ We know that, $T=2 \pi \sqrt{\;\frac{l}{g}}$ $\Rightarrow \;\; T^{2}=4 \pi ^{2} \;\frac{I}{g} \Rightarrow g=\;\frac{4 \pi ^{2} I}{T^{2}}$ $\Rightarrow \;\; \;\frac{\Delta g}{g}=\;\frac{\Delta l}{l}+2 \;\frac{\Delta T}{T}=\;\frac{0.1}{10}+2 \;\frac{1}{100}$ $=0.01+0.02=0.03$ $\Rightarrow \;\; \;\frac{\Delta g}{g} \times 100=0.03 \times 100$ $=3 \%$

Given, mass of particle, m = 1kg Spring constant, $k=100 {\text{Nm}}^{-1}$ Let time period is T. Kinetic energy (KE) = Potential energy (PE) = E Amplitude = A Angular velocity = ω Wave displacement = x KE in SHM = PE in SHM $\Rightarrow \frac{1}{2} m \omega ^{2}(\text{A}^{2}-x^{2})=\frac{1}{2} k x^{2}$ $\Rightarrow m \omega ^{2}(\text{A}^{2}-x^{2})=k x^{2}$ ...(i) Since, force, $\text{F}=m \omega ^{2} x=k x$ $\Rightarrow k=m \omega ^{2}$ Substituting the value in Eq. (i), we get $\Rightarrow m \omega ^{2}(\text{A}^{2}-x^{2})=m \omega ^{2} x^{2}$ $\Rightarrow \text{A}^{2}-x^{2}=x^{2}$ $\Rightarrow \text{A}^{2}=2 x^{2}$ $\Rightarrow x=\text{A} / \\sqrt{2}$ Since, wave displacement, $x=\text{A} \sin \omega t$ $\Rightarrow \frac{\text{A}}{\\sqrt{2}}=\text{A} \sin \omega t$ $\Rightarrow \sin (\pi / 4)=\sin \omega t$ $\Rightarrow \omega t=\pi / 4$ $\Rightarrow \frac{2 \pi }{\text{T}} .t=\frac{\pi }{4}$ $\Rightarrow t=\frac{\text{T}}{8}$ Comparing the given value in the question i.e. $\frac{\text{T}}{x}$, the value of x = 8

Given, radius of earth $=R$ Distance of chord from centre of earth $=\;\frac{R}{2}$ Let $x_{1}$ be the radius of inner circle and $M$ be the mass of earth. $\therefore m^{'}(\;\text{ effective mass of earth }\;) \;\; =\;\frac{M}{\frac{4}{3} \pi R^{3}} \cdot \;\frac{4}{3} \pi x_{1}^{3}$ $\Rightarrow \;\; m^{'} \;\; =\;\frac{M}{R^{3}} x_{1}^{3}$ If $F$ is the gravitational force exerted by earth on particle at position $x$ and $\omega$ be the angular velocity in time period $T$, then $F \;\; =\;\frac{G m^{'} m}{x_{1}^{2}}=\;\frac{G m}{x_{1}^{2}} \cdot \;\frac{M}{R^{3}} x_{1}^{3}$ $\Rightarrow \;\; m \omega ^{2} x_{1} \;\; =\;\frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \;\; \omega \;\; =\sqrt{\;\frac{G M}{R^{3}}}$ $F=\;\frac{G m^{'} m}{x_{1}^{2}}=\;\frac{G m}{x_{1}^{2}} \cdot R^{3} x_{1}^{3}$ $\Rightarrow \;\; m \omega ^{2} x_{1}=\;\frac{G M m x_{1}}{R^{3}}$ $\Rightarrow \;\; \omega =\sqrt{\;\frac{G M}{R^{3}}}$ Since, $\omega =\;\frac{2 \pi }{T}$ and $G M=g R^{2}$ Substituting the above values in Eq. (i), we get 1$\Rightarrow \;\; \;\frac{2 \pi }{T}=\sqrt{\;\frac{g R^{2}}{R^{3}}} \Rightarrow T=2 \pi \sqrt{\;\frac{R}{g}}$

Given, the time period of the simple harmonic motion, $T=2 {s}$ Displacement covered by the particle, $x=A / 2$ Here, $A$ is the amplitude of the particle. The general equation of the simple harmonic motion, $x=A \sin \omega t$ $\Rightarrow \;\; \;\frac{A}{2}=A \sin (\;\frac{2 \pi }{T} t) \Rightarrow \;\frac{1}{2}=\sin (\;\frac{2 \pi }{2} t)$ $\Rightarrow \;\; \sin \;\frac{\pi }{6}=\sin (\pi t) \Rightarrow \;\frac{\pi }{6}=\pi t$ $\Rightarrow \;\; t=\;\frac{1}{6}$ The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $1/6 {s}$. Comparing with $t=1/a$ Hence, the value of the $a$ is 6 .

We know that, the expression of maximum velocity during oscillation, $V_{\max }=A \omega$ Given, $V_{\max }(A)=V_{\max }(B) \Rightarrow A_{1} \omega _{1}=A_{2} \omega _{2}$ Now, the ratio of the amplitude during oscillation, $\;\frac{A_{1}}{A_{2}}=\;\frac{\omega _{2}}{\omega _{1}}$ We know that, $\omega =\sqrt{k / m}$ where, $k$ is the spring constant, $m$ is the mass of the object. Substituting the value of $\omega$ in Eq. (i), we get $\;\frac{A_{1}}{A_{2}}=\sqrt{\;\frac{k_{2}}{k_{1}}}$

Potential energy of ${SHM}=\;\frac{1}{2} k x^{2}=\;\frac{1}{2} m \omega ^{2} x^{2}$ where $m=$ mass of particle, $\omega =$ angular velocity and $x=$ displacement. Kinetic energy of ${SHM}=\;\frac{1}{2} m(\omega \sqrt{A^{2}-x^{2}})^{2}$ Here, $A$ is the amplitude of SHM. According to question, Potential energy of ${SHM}=$ Kinetic energy of ${SHM}$ $\;\frac{1}{2} m \omega ^{2} x^{2}=\;\frac{1}{2} m(\omega \sqrt{A^{2}-x^{2}})^{2}$ $\Rightarrow \;\; \omega ^{2} x^{2}=\omega ^{2}(\sqrt{A^{2}-x^{2}})^{2} \Rightarrow \omega ^{2} x^{2}=\omega ^{2}(A^{2}-x^{2})$ $\Rightarrow \;\; \omega ^{2} x^{2}=\omega ^{2} A^{2}-\omega ^{2} x^{2} \Rightarrow 2 \omega ^{2} x^{2}=\omega ^{2} A^{2}$ $\Rightarrow \;\; 2 x^{2}=A^{2} \Rightarrow x^{2}=\;\frac{A^{2}}{2}$ $\Rightarrow \;\; x=± \sqrt{\;\frac{A^{2}}{2}} \Rightarrow x=± \;\frac{A}{\sqrt{2}}$

${v}^{2}=\omega ^{2}({A}^{2}-{x}^{2})$ ${A}^{2}={x}_{1}^{2}+\frac{{v}_{1}^{2}}{\omega ^{2}}={x}_{2}^{2}+\frac{{v}_{2}^{2}}{\omega ^{2}}$ $\omega ^{2}=\frac{{v}_{2}^{2}-{v}_{1}^{2}}{{x}_{1}^{2}-{x}_{2}^{2}}$ ${T}=2 \pi \\sqrt{\frac{{x}_{2}^{2}-{x}_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

${E}=\frac{1}{2} {Ka}^{2}$ $\frac{3 {E}}{4}=\frac{1}{2} {K}({a}^{2}-{y}^{2})$ $\frac{3}{4} \\times \frac{1}{2} {Ka}^{2}=\frac{1}{2} {K}({a}^{2}-{y}^{2})$ ${y}^{2}={a}^{2}-\frac{3 {a}^{2}}{4}$ ${y}=\frac{{a}}{2}$

Given, displacement-time equation, $Y=A \sin (\omega t+\varphi _{0})$ Here, $A$ is amplitude, $\omega$ is angular frequency, $t$ is time taken and $\varphi _{0}$ is the phase constant. At $t=0, Y=A / 2$ $\therefore \;Y=A / 2=A \sin (0+\varphi _{0})=A \sin \varphi _{0}$ $\Rightarrow \;\sin \varphi _{0}=1 / 2$ $\Rightarrow \;\varphi _{0}=\pi / 6$

Time period of Fig. 1 can be given as $T_{a}=2 \pi \sqrt{\;\frac{M}{k}}$ where, $M$ is mass of the suspended object and $k$ is the force constant. In Fig. 2, both the springs are in series combination, therefore its time period can be given as $T_{b}=2 \pi \sqrt{\;{M}/{k_{ {eq}}}}=2 \pi \sqrt{\;\frac{M}{k {/} 2}} \;\; ( \because k_{ {eq}}=\;\frac{k \times k}{k+k})$ Now, $\;\frac{T_{b}}{T_{a}}=\;{2 \pi \sqrt{\frac{M}{k {/} 2}}}/{2 \pi \sqrt{\;\frac{M}{k}}}$ $\Rightarrow \;\; \;\frac{T_{b}}{T_{a}}=\sqrt{2}$...(i) According to question, the ratio of time period of oscillation of two ${SHM}$ is $T_{b} {/} T_{a}=\sqrt{x}$, so on comparing it with Eq. (i) we can say, $x=2$

Time period, $T=\;\frac{2 \pi }{\omega ^{'}}$ Given, time period, $T=\;\frac{\pi }{\omega }$ $\;\frac{2 \pi }{\omega ^{'}}=\;\frac{\pi }{\omega } \Rightarrow \omega ^{'}=2 \omega$ Options (a) and (b) are incorrect. Option (c), $\sin ^{2} \omega t=\;\frac{1}{2}(2 \sin ^{2} \omega t)=\;\frac{1}{2}(1-\cos 2 \omega t)$ Hence, the angular frequency is $2 \omega$. Option (d), $3 \cos (\;\frac{\pi }{4}-2 \omega t)$ Angular frequency of ${SHM}$ is $2 \omega$. Option (d) is the correct answer.

${T}=2 \pi \\sqrt{\frac{{m}}{{k}}}$ $0.2=2 \pi \\sqrt{\frac{0.5}{{k}}}$ ${k}=50 \pi ^{2}$ $\approx 500$ ${x}={A} \sin (\omega {t}+\phi )$ $=5 {cm} \sin (\frac{\omega {T}}{4}+0)$ $=5 {cm} \sin (\frac{\pi }{2})$ $=5 {cm}$ ${PE}=\frac{1}{2} {kx}^{2}$ $=\frac{1}{2}(500)(\frac{5}{100})^{2}$ $=0.6255$

Let spring constants of two springs be $k_{1}$ and $k_{2} .$ Since, two springs are connected in parallel connection and parallel equivalent spring constant, $k_{\;\text{eq }\;}=k_{1}+k_{2}$ $\Rightarrow \;\; k_{ {eq}}=2 k+2 k=4 k$ As, time period, $T=2 \pi \sqrt{\;{m}/{k_{\;\text{eq }\;}}}$ $\Rightarrow \;\; T=2 \pi \sqrt{\;\frac{m}{4 k}}=\;\frac{2 \pi }{2} \sqrt{\;\frac{m}{k}}=\pi \sqrt{\;\frac{m}{k}}$

Given, $y_{1}=10 \sin (3 \pi t+\;\frac{\pi }{3})$ $A_{1} \;\; =10$ $y_{2} \;\; =5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)$ $\;\; =5 \sin 3 \pi t+5 \sqrt{3} \cos 3 \pi t$ $\;\text{ Amplitude of }\; {SHM}_{2} \;\text{ is }\;$ ${A}_{2} \;\; =\sqrt{5^{2}+(5 \sqrt{3})^{2}}$ $\;\; =\sqrt{25+75}$ $\;\; =\sqrt{100}=10$ $\therefore \;\text{ Ratio of amplitudes, }\; \;\frac{A_{1}}{A_{2}}=\;\frac{10}{10}=1: 1$ $\;\; x=1$