Oscillations Practice Questions

The amplitude of the damped harmonic oscillator is given by the envelope function $A(t) = A_0 e^{-0.1t}$, where $A_0$ is the initial amplitude. We seek the time $t$ such that $A(t) = \frac{A_0}{2}$.
Substituting this condition gives: $\frac{A_0}{2} = A_0 e^{-0.1t} \implies \frac{1}{2} = e^{-0.1t}$.
Inverting both sides, we get: $2 = e^{0.1t}$.
Taking the natural logarithm: $\ln(2) = 0.1t \implies t = \frac{\ln(2)}{0.1}$.
Using $\ln(2) \approx 0.693$, we find $t = 10 \times 0.693 = 6.93 \text{ s}$.
Rounding to the nearest value, we get $t \approx 7 \text{ s}$.

τ = - 2K × $l/2$ cos θ \Rightarrow τ = $({Kl^2}/2)$ θ = - C θ \Rightarrow f = $1/{2\pi }$ $\sqrt{C/I}$ = $1/{2\pi }$ $\sqrt{\frac{{Kl^2}/2}{\frac{Ml^2}{12}}}$ \Rightarrow f = $1/{2\pi }$ $\sqrt{{6k}/m}$

Maximum kinetic energy at lowest point B is given by K = mgl (1 - cos θ) where θ = angular amp. $K_1$ = mg l (1 - cos θ) $K_2$ = mg(2 l) (1 - cos θ) $K_2$ = $2K_1$.

Extra Boyant force = δAxg $B_0$ + B × mg = ma B = ma a = $({\delta Ag}/m)^x$ $w^2$ = ${\delta Ag}/m$ w = $\sqrt{\frac{10^3\times \pi (2.5)^2\times 10^{-4}\times 10}{310\times 10^{-6}\times 10^3}}$ = $\sqrt{63.30}$ = 7.95

Potential energy (U) = $1/2 kx^2$ Kinetic energy (K) = $1/2 kA^2$ - $1/2kx^2$ According to the question, U = k ∴ $1/2 kx^2$ = $1/2kA^{2-}1/2kx^2$ x = ± $A/\sqrt2$ ∴ Correct answer is (3)

Angular frequency of pendulum ω = $\sqrt{g_{\text{eff}}/l}$ ∴ ${\Delta \omega }/\omega$ = $1/2 {\Delta g_{\text{eff}}}/g_{\text{eff}}$ Δω = $1/2 {\Delta g}/g$ × ω [$\omega _s$ = angular frequency of support] Δω = $1/2 \times \frac{2A\omega _s^2}{100}$ × 100 Δω = $10^{-3}$ rad/s

$T= \frac{1}{5} \sec$ $t_{1 / 2}=2 {sec}$for $\frac{A}{2^{10}}$ approx. time $=20 {sec}$

The equation for simple harmonic motion is given by $x(t) = A \cos(\omega t + \phi)$. Given $x(t_0) = a$, $x(2t_0) = b$, and $x(3t_0) = c$, let $\theta = \omega t_0$. Then, $a = A \cos(\phi + \theta)$, $b = A \cos(\phi + 2\theta)$, and $c = A \cos(\phi + 3\theta)$.
Summing $a$ and $c$, we have $a + c = A [\cos(\phi + \theta) + \cos(\phi + 3\theta)]$.
Using the trigonometric identity $\cos(\alpha) + \cos(\beta) = 2 \cos(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})$, we get:
$a + c = A \cdot 2 \cos\left(\frac{2\phi + 4\theta}{2}\right) \cos\left(\frac{-2\theta}{2}\right) = 2 A \cos(\phi + 2\theta) \cos(\theta)$
Since $b = A \cos(\phi + 2\theta)$, this simplifies to $a + c = 2b \cos(\theta)$, which implies $\cos(\omega t_0) = \frac{a+c}{2b}$.
Solving for frequency $f = \frac{\omega}{2\pi}$, we find $\omega t_0 = \cos^{-1}\left(\frac{a+c}{2b}\right)$, leading to $f = \frac{1}{2\pi t_0} \cos^{-1}\left(\frac{a+c}{2b}\right)$.

Momentum conservation at each equilibrium crossing indicates that for even-numbered collisions, the velocity $v=0$ because the oscillator's momentum cancels the incoming particle's momentum. For odd-numbered collisions $n$, the effective velocity is $v_n = \frac{m u}{M+n m}$. The amplitude is $A_n = v_n \sqrt{\frac{M+n m}{k}} = \frac{m u}{\sqrt{k(M+n m)}}$.

Substituting $M=10$, $m=5$, $u=1$, $k=1$, and $n=13$:
$A_{13} = \frac{5 \cdot 1}{\sqrt{1 \cdot (10 + 13 \cdot 5)}} = \frac{5}{\sqrt{10 + 65}} = \frac{5}{\sqrt{75}} = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}}$

Given the equations $x(t) = A \sin(at + \delta)$ and $y(t) = B \sin(bt)$, we evaluate the parameter set (4).
For $a=b$ and $\delta=\pi/2$, the equations become $x = A \sin(bt + \pi/2) = A \cos(bt)$ and $y = B \sin(bt)$.
Rearranging, we find $\frac{x}{A} = \cos(bt)$ and $\frac{y}{B} = \sin(bt)$.
Using the identity $\cos^2(bt) + \sin^2(bt) = 1$, we get $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.
Since $A \neq B$, this equation represents the standard form of an ellipse.
Thus, the parameters in (4) correctly describe an ellipse.

$w=\sqrt{k/m}$ $2\times \pi \times 10^12=\sqrt{k/108/{6\times 10^23}}$k nearly 7.1 N/m

The mechanical energy $E(t)$ of a damped harmonic oscillator decays as $E(t) = E_0 e^{-(b/m)t}$. Setting $E(t) = \frac{E_0}{2}$ to find the half-life of the energy, we obtain $\frac{1}{2} = e^{-(b/m)t}$, which simplifies to $t = \frac{m \ln(2)}{b}$. Substituting the given values $m = 0.1 \, \text{kg}$ and $b = 10^{-2} \, \text{kg} \, \text{s}^{-1}$:

$t = \frac{0.1 \, \text{kg} \times \ln(2)}{10^{-2} \, \text{kg} \, \text{s}^{-1}} = 10 \times 0.693 \approx 6.93 \, \text{s}$

This value is closest to $7 \, \text{s}$.

$\frac{\omega a}{a\omega }=10$ $w=10$ $x=asin(\omega +\pi /4)$at $f=0$ $5=asin(\pi /4)$ $a=5\sqrt2$max acc. $=w2a$ $=100\times 5\sqrt2$ $=500\sqrt2$

Time taken to reach the extreme position from equilibrium position is $T/4$.Velocity is maximum at equilibrium position and zero at extreme position.$V=A\;\omega \;cos\;\omega t$ K.E$=1/2mv^2$ (m is the mass of particle and v is the velocity of particle)K.E. $=1/2m A^2\omega ^2cos^2\omega t$Hence graph of K.E. v/s time is square cos function

$F=1/{2\pi }\sqrt{k/m}=1$ $4\pi ^2=k/m$ $m=1$ $k=4\pi ^2$In parallel $k_\text{eq}=2k$ $F=1/{2\pi }\sqrt{{K\times 2}/8}$ $=1/2$Hz

$\omega ^2A=g$ $T={2\pi }/\omega$ $\omega =2\pi t$ $\omega ^2(0.07)=10m/s^2$ $4\pi ^2f^2(0.07)=10m/s^2$ $f=5/\sqrt7Hz=1.9Hz$

$t=\pi /{3\omega }=T/6$

$3\omega \sqrt{A^{2-}({2A}/3)^2}=\omega \sqrt{A_1^{2-}({2A}/3)^2}$ $\therefore A_1={7A}/3$