Oscillations Practice Questions

As we know that, for a body executing SHM the average kinetic energy is equal to average potential energy over a complete time period i.e. ${\text{KE}}_{{\text{av}}}={\text{PE}}_{{\text{av}}}=\frac{1}{4} m \omega ^{2} {a}^{2}$ ∴ Total mechanical energy at any time = Sum of kinetic and potential energy $=\frac{1}{2} m \omega ^{2} a^{2}=$ constant. where, m = mass of body executing SHM, ω = angular frequency and a = amplitude of SHM. Hence, (c) and (d) are true.

Statement $I$ is false because time period of second's pendulum is always $2 {s}$. Therefore, time taken to move between two extreme positions will be $T / 2=2 / 2=1 {s}$ Hence, option (b) is the correct.

$K=\frac{1}{2} m \omega ^{2}(A^{2}-x^{2})$ $=\frac{1}{2} m \omega ^{2}(A^{2}-\frac{A^{2}}{4})$ $=\frac{1}{2} m \omega ^{2}(\frac{3 A^{2}}{4})$ $K=\frac{3}{4}(\frac{1}{2} m \omega ^{2} A^{2})$

${T}_{0}=2 \pi \\sqrt{\frac{l}{{g}}}$ New time period ${T}=2 \pi \\sqrt{\frac{l/ 16}{{ g}}}=\frac{2 \pi }{4} \\sqrt{\frac{l}{{g}}}$ ${T}=\frac{{T}_{0}}{4}$

Since, the particle is executing SHM. Therefore, displacement equation of wave will be $y \;\; =A \sin \omega t$ $\Rightarrow \;\; y / A \;\; =\sin \omega t$ and wave velocity equation will be $\;\;\;\;\; v_{y}=\;\frac{d y}{d t}=A \omega \cos \omega t$ $\Rightarrow \;\; v_{y} / A \omega =\cos \omega t$ $\;\text{ Now, }\; \;\; \sin ^{2} \omega t+\cos ^{2} \omega t=1$ $\therefore \;\; (y / A)^{2}+(v_{y} / A \omega )^{2}=1$ This equation is similar to the equation of ellipse.

Given, angular displacement to complete $\;\frac{5}{8}(=\;\frac{1}{2}+\;\frac{1}{8})$ rev $\;\; =(\pi +\;\frac{\pi }{6}) {rad}$ $\;\; =(\;\frac{7 \pi }{6}) {rad}$ Since, $\;\; \omega =\;\frac{2 \pi }{T}=\;\frac{\theta }{t}$ $\Rightarrow \;\; \;\; \theta \;=\;\frac{2 \pi }{T} \cdot t \Rightarrow \;\frac{7 \pi }{6}=\;\frac{2 \pi }{T} t$ $\Rightarrow \;\; t\;=\;\frac{7 T}{12}=\;\frac{\alpha }{\beta } T$ Hence, $\alpha =7$

Time period of a simple pendulum can be given as $T=2 \pi \sqrt{\;\frac{I}{g}}$...(i) When the lift moves upwards, then effective acceleration is $\Rightarrow$ $\Rightarrow g_{\;\text{eff }\;}=g+a=g+\;\frac{g}{2}=\;\frac{3 g}{2}$ $\therefore$ New time period, $T_{1}=2 \pi \sqrt{\;{1}/{g_{\;\text{eff }\;}}}=2 \pi \sqrt{\;\frac{2 L}{3 g}}$ $\Rightarrow \;\; T_{1}=\sqrt{\;\frac{2}{3}} T$ [From Eq. (i)]

It is given that motion is free simple harmonic motion. Consider a spring mass system performing free simple harmonic motion. Displacement of mass is given as $x=x_{0} \sin \omega t$ The potential energy stored in spring for displacement x can be given as ${\text{PE}}=\frac{1}{2} k x^{2}$ Substituting the value of x, we get ${\text{PE}}=\frac{1}{2} k x_{0}^{2} \sin ^{2} \omega t$ Here, k is spring constant and $x_0$ is also constant. $\text{PE}=\text{C} \sin ^{2} \omega t$ $(\therefore \text{C}=\frac{1}{2} k x_{0}^{2})$ Simplify the value of $\sin^2$ ωt, we get $\text{PE}=\text{C}[\frac{1-\cos2\omega \text{t}}{2}]$ The value of function cos2ωt will always lies between −1 and 1. Thus, the value of potential energy will be greater than or equal to zero. 0 ≤ PE This condition is only satisfies by graph (d).

${x}({t})={A} \sin (\omega {t}+\phi )$ $v(t)=A \omega \cos (\omega t+\phi )$ $2={A} \sin \phi$ ........(1) $2 \omega =A \omega \cos \phi$ .......(2) From (1) and (2) $\tan \phi =1$ $\phi =45°$ Putting value of $\phi$ in equation (1) $2={A}\{\frac{1}{\\sqrt{2}}\}$ ${A}=2 \\sqrt{2}$ ${x}=2$

$T=2\pi \sqrt{l/g}$ $g={4\pi ^2l}/T^2$ ${\Deltag}/g={\Deltal}/l+{2\DeltaT}/T$ $=0.1/25+{2\times 1}/50=4.4$

${ (A) } {F}= {ma}\;\;\;\ {a}=-\omega ^{2} {x}$ at $\frac{3 {T}}{4}$ displacement zero $( {x}=0),$ so ${a}=0$ ${F}=0$ (B) at ${t}={T} \;\;\;$ displacement $( {x})= {A}$ x maximum, So acceleration is maximum. (C) ${V}=\omega \sqrt{ {A}^{2}- {x}^{2}}$ ${V}_{\max }$ at ${x}=0$ ${V}_{\max }={A} \omega$ at ${t}=\frac{{T}}{4}, \ {x}=0,$ So ${V}_{ {\max}}$ (D) ${KE}={PE}$ $\therefore$ at $x=\ \frac{A}{\\sqrt{2}}$ at $t=\ \frac{T}{2}\;\; x=-A \;\;$ (So not possible)

At equilibrium position ${V}_{0}=\omega _{0} {A}=\sqrt{\frac{{K}}{ {m}}} {A}$ ...(i) ${V}=\omega {A}^{1}=\sqrt{\frac{{K}}{ {m}}} {A}^{1}$ .....(ii) $\therefore \ A^{1}= \frac{A}{\\sqrt{2}}$

FBD of m in frame of disc/- $k\Deltal=m\omega ^2(l_0+\Deltal)$ $\Deltal=\frac{m\omega ^2l_0}{k-m\omega ^2}\approx {m\omega l_0}/k$ ${\Deltal}/l_0=$ Relative change $={m\omega ^2}/k$

${y}={y}_{0} \sin ^{2} \omega {t}$ ${y}=\frac{{y}_{0}}{2}(1-\cos 2 \omega {t})$ ${y}-\frac{{y}_{0}}{2}=-\frac{{y}_{0}}{2} \cos 2 \omega {t}$ Amplitude : $\frac{{y}_{0}}{2}$ $\frac{{y}_{0}}{2}=\frac{{mg}}{{K}}$ $2 \omega =\sqrt{\frac{{K}}{{m}}}=\sqrt{\frac{2 {g}}{{y}_{0}}}$ $\omega =\sqrt{\frac{{g}}{2 {y}_{0}}}$

k = $1/2 m\omega ^2A^2cos^2\omega t$ U = $1/2 m\omega ^2A^2sin^2 \omega t$ $k/U$ = $cot^2$ ωt = $cot^2$ $\pi /{90}$ (210) = $1/3$ Hence ratio is 3 (most appropriate)

Since g = ${GM}/R^2$ $g_p/g_e$ = $M_e/M_e (R_e/R_p)^2$ = 3 $(1/3)^2$ = $1/3$ Also T α $1/\text{sqrtg}$ \Rightarrow $T_p/T_e$ = $\sqrt{g_e/g_p}$ = $\sqrt3$ \Rightarrow $T_p$ = $2\sqrt3$ s

For closed organ pipe, resonate frequency is odd multiple of fundamental frequency. ∴ (2n + 1) $f_0$ ≤ 20,000 ($f_o$ is fundamental frequency = 1.5 KHz) ∴ n = 6 ∴ Total number of overtone that can be heared is 7. (0 to 6).

y = 5(sin3πt+ $\sqrt3$ cos3πt) cm = 10 sin $(3\pi t+\pi /3)$ Amplitude = 10 cm T = ${2\pi }/w$ = $\frac{2\pi }{3\pi }$ = $2/3$ s