Different types of oscillations (Free, Damped, Forced Oscillation and Resonance) Questions in English

(D) In forced oscillations,the amplitude of vibration at resonance is given by $A = \frac{F_0}{\sqrt{m^2(\omega^2 - \omega_0^2)^2 + b^2\omega^2}}$,where $b$ is the damping constant.
When the damping force is small (i.e.,$b$ is small),the denominator becomes very small at resonance (where $\omega \approx \omega_0$),leading to a very large amplitude.
Furthermore,the sharpness of the resonance curve is inversely proportional to the damping. $A$ smaller damping force results in a narrower and taller resonance peak,making the resonance curve sharper.

(B) Resonance occurs when the amplitude of the oscillation becomes infinite $(A \to \infty)$.
Given the expression for amplitude: $A = \frac{c}{a + b - c}$.
For $A$ to be infinite,the denominator must be zero:
$a + b - c = 0$.
If we set $b = 0$,the condition becomes $a - c = 0$,which implies $a = c$.
Therefore,resonance occurs when $b = 0$ and $a = c$.

(B) For a damped driven oscillator,the amplitude $A$ is given by $A = \frac{F/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (b\omega/m)^2}}$.
Amplitude is maximum when the denominator is minimum,which occurs at $\omega_1 = \sqrt{\omega_0^2 - 2(b/2m)^2}$. Thus,$\omega_1 \neq \omega_0$.
The energy of the oscillator is proportional to the square of the amplitude and is also related to the power absorbed. The energy of the particle is maximum at the velocity resonance frequency,which occurs when $\omega_2 = \omega_0$.
Therefore,the correct relationship is $\omega_1 \neq \omega_0$ and $\omega_2 = \omega_0$.

(A) simple pendulum oscillating in a real-world environment experiences resistive forces,primarily air resistance (or air friction).
These resistive forces act in the direction opposite to the velocity of the bob.
This causes the mechanical energy of the system to dissipate as heat,leading to damped oscillations.
As the amplitude of the oscillation gradually decreases over time,the pendulum eventually comes to rest.

(C) simple pendulum oscillating in air experiences air resistance (damping force).
The damping force acts opposite to the direction of motion,which causes the energy of the oscillator to dissipate over time.
As a result,the amplitude $A$ of the oscillation gradually decreases.
For a damped harmonic oscillator,the time period $T'$ is given by $T' = \frac{2\pi}{\sqrt{\omega_0^2 - b^2/4m^2}}$,where $b$ is the damping constant. As damping occurs,the effective frequency decreases,which means the time period $T$ increases.
Therefore,as time passes,$T$ increases and $A$ decreases.

(A) For resonance,the amplitude $a_m$ must be maximum.
This occurs when the denominator $f(\omega) = a\omega^2 - b\omega + c$ is at its minimum value.
For a single resonant frequency,the quadratic equation $a\omega^2 - b\omega + c = 0$ must have exactly one real root for $\omega$.
This happens when the discriminant $D = b^2 - 4ac$ is equal to zero.
Therefore,the condition for a single resonant frequency is $b^2 = 4ac$.

(B) For a forced oscillator,the equation of motion is given by $m \frac{d^2x}{dt^2} + kx = F_0 \cos \omega t$.
Substituting $x = x_0 \cos \omega t$,we get $-m \omega^2 x_0 + k x_0 = F_0$.
Since the natural angular frequency is $\omega_0 = \sqrt{k/m}$,we have $k = m \omega_0^2$.
Substituting this,we get $m x_0 (\omega_0^2 - \omega^2) = F_0$.
Thus,the amplitude $x_0 = \frac{F_0}{m(\omega_0^2 - \omega^2)}$.
Therefore,the displacement $x$ is proportional to $\frac{1}{m(\omega_0^2 - \omega^2)}$.

(B) The amplitude of a damped oscillator is given by the equation $A = A_0 e^{-\lambda t}$,where $A_0$ is the initial amplitude,$\lambda$ is the damping constant,and $t$ is time.
Given that at $t = 1 \text{ min}$,the amplitude becomes half:
$\frac{A_0}{2} = A_0 e^{-\lambda(1)}$
$\frac{1}{2} = e^{-\lambda} \Rightarrow e^{\lambda} = 2$.
Now,for $t = 3 \text{ min}$,the amplitude $A$ is:
$A = A_0 e^{-\lambda(3)} = A_0 (e^{-\lambda})^3 = A_0 \left(\frac{1}{e^{\lambda}}\right)^3$
Substituting $e^{\lambda} = 2$:
$A = A_0 \left(\frac{1}{2}\right)^3 = \frac{A_0}{2^3}$.
Comparing this to $\frac{A_0}{X}$,we get $X = 2^3$.

(A) The equation of motion for a damped pendulum is given by $I \alpha = -mg \ell \theta - b' v \ell$,where $b'$ is the damping constant. For a spherical bob,the drag force is $F_d = -bv$. The angular displacement follows the form $\theta(t) = \theta_0 e^{-(b/2m)t} \sin(\omega t + \phi)$.
Given the amplitude decays as $A(t) = \theta_0 e^{-(b/2m)t}$,we define the average life $\tau$ as the time when the amplitude becomes $1/e$ of the initial amplitude $\theta_0$.
Thus,$\theta_0/e = \theta_0 e^{-(b/2m)\tau}$.
Comparing the exponents,we get $1 = (b/2m)\tau$.
Assuming the proportionality constant $b$ in the question refers to the damping factor per unit mass (often denoted as $b/m$),the average life is $\tau = 2/b$.

(D) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-bt/2m}$.
Given that at $t = 5 \ s$,$A = 0.9 A_0$. Substituting this into the equation:
$0.9 A_0 = A_0 e^{-b(5)/2m} \implies e^{-5b/2m} = 0.9$.
We need to find the amplitude after another $10 \ s$,which means at total time $t = 5 + 10 = 15 \ s$.
$A(15) = A_0 e^{-b(15)/2m} = A_0 (e^{-5b/2m})^3$.
Substituting the value $e^{-5b/2m} = 0.9$:
$A(15) = A_0 (0.9)^3 = A_0 (0.729)$.
Thus,$\alpha = 0.729$.

(A) $1$. Calculate the damping constant $b$: Since the force $F = bv$,we have $b = F/v = 120/1 = 120 \ kg/s$.
$2$. Calculate the damping parameter $r$: $r = b / (2m) = 120 / (2 \times 700) = 120 / 1400 = 6/70 \approx 0.0857 \ s^{-1}$.
$3$. Calculate the spring constant $k$: From $F = kx$,we have $k = 450 / 2 = 225 \ N/m$.
$4$. Calculate the natural angular frequency $\omega_0$: $\omega_0 = \sqrt{k/m} = \sqrt{225/700} = \sqrt{9/28} \approx 0.5669 \ rad/s$.
$5$. Calculate the angular frequency of damped $SHM$ $\omega'$: $\omega' = \sqrt{\omega_0^2 - r^2} = \sqrt{225/700 - (6/70)^2} = \sqrt{22500/70000 - 36/4900} = \sqrt{225/700 - 36/4900} = \sqrt{(1575 - 36)/4900} = \sqrt{1539}/70 \approx 39.23 / 70 \approx 0.56 \ rad/s$.

(C) The amplitude of a damped harmonic oscillator is given by $A(t) = A_0 e^{-bt}$.
Given that the amplitude becomes half in $3 \ s$:
$A(3) = A_0 e^{-3b} = \frac{A_0}{2}$
$e^{-3b} = \frac{1}{2} \quad \dots (i)$
In the next $6 \ s$,the total time elapsed is $3 \ s + 6 \ s = 9 \ s$. The amplitude at $t = 9 \ s$ is:
$A(9) = A_0 e^{-9b} = A_0 (e^{-3b})^3$
Substituting equation $(i)$ into the expression:
$A(9) = A_0 \left(\frac{1}{2}\right)^3 = A_0 \left(\frac{1}{8}\right)$
Since the amplitude becomes $1/x$ of the initial amplitude,we have $1/x = 1/8$,which means $x = 8 = 2^3$.
Therefore,the correct option is $C$.

(C) When a pendulum oscillates in a medium like water,it experiences a damping force due to the viscosity of the medium. This damping force causes the amplitude of oscillation to decrease exponentially over time. The total mechanical energy $E$ of a damped harmonic oscillator is proportional to the square of its amplitude $A$. Since the amplitude $A$ decays exponentially as $A(t) = A_0 e^{-\gamma t}$,the total mechanical energy $E$ also decays exponentially with time according to the relation $E(t) = E_0 e^{-2\gamma t}$. This exponential decay is represented by a curve that starts at a maximum value and approaches zero asymptotically as time increases. Among the given options,diagram $C$ correctly depicts this exponential decay of energy over time.

(B) At terminal velocity,the drag force equals the gravitational force: $mg = bv_{t}$.
Given $m = 3 \ kg$,$v_{t} = 25 \ m/s$,and taking $g = 10 \ m/s^2$,we have $3 \times 10 = b \times 25$,which gives $b = \frac{30}{25} = 1.2 \ kg/s$.
The damping constant $\gamma = \frac{b}{2m} = \frac{1.2}{2 \times 3} = 0.2 \ s^{-1}$.
The angular frequency of damped $SHM$ is given by $\omega' = \sqrt{\omega_0^2 - \gamma^2}$,where $\omega_0 = \sqrt{\frac{k}{m}}$.
$\omega_0 = \sqrt{\frac{300}{3}} = \sqrt{100} = 10 \ rad/s$.
$\omega' = \sqrt{10^2 - (0.2)^2} = \sqrt{100 - 0.04} = \sqrt{99.96}$.
Using the binomial approximation $\sqrt{x^2 - a} \approx x - \frac{a}{2x}$ for small $a$,we get $\omega' \approx 10 - \frac{0.04}{20} = 10 - 0.002 = 9.998 \ rad/s$.

(B) The amplitude of a damped oscillator decays with a time constant $\tau = \frac{2m}{b}$.
Given that both systems have identical decay times,$\tau_A = \tau_B$.
This implies $\frac{2m_A}{b_A} = \frac{2m_B}{b_B}$.
Since $m_A = 2m_B$,we substitute this to get $\frac{2(2m_B)}{b_A} = \frac{2m_B}{b_B}$.
Simplifying this equation,we get $b_A = 2b_B$.

(C) When a pendulum oscillates in a medium like water,it experiences a resistive force (damping force) proportional to its velocity. This leads to damped oscillations. The total mechanical energy $E$ of a damped oscillator at any time $t$ is given by the expression: $E = E_0 e^{-\gamma t}$,where $E_0$ is the initial energy and $\gamma$ is the damping constant. This equation represents an exponential decay of energy with time. Among the given options,the graph that shows an exponential decrease in energy $E$ as time $t$ increases is represented by graph $C$.

(D) The angular frequency of damped oscillations is given by $\omega' = \sqrt{\frac{K}{m} - \frac{b^2}{4m^2}}$.
Given $T = \frac{\pi}{12} \ s$,the angular frequency is $\omega' = \frac{2\pi}{T} = \frac{2\pi}{\pi/12} = 24 \ rad/s$.
Substitute the values $K = 1250 \ N/m$,$m = 2 \ kg$,and $\omega' = 24 \ rad/s$ into the equation:
$24 = \sqrt{\frac{1250}{2} - \frac{b^2}{4(2^2)}}$
$24 = \sqrt{625 - \frac{b^2}{16}}$
Squaring both sides:
$576 = 625 - \frac{b^2}{16}$
$\frac{b^2}{16} = 625 - 576 = 49$
$b^2 = 49 \times 16 = 784$
$b = \sqrt{784} = 28 \ kg/s$.

(B) In damped oscillations,the amplitude decays exponentially with time,given by $a = a_0 e^{-bt}$,where $b$ is the damping constant and $t$ is the time.
Let $T$ be the time period of one oscillation. After $n$ oscillations,the time elapsed is $t = nT$.
Initially,after $100$ oscillations,the amplitude is $a = \frac{a_0}{3}$.
So,$\frac{a_0}{3} = a_0 e^{-b(100T)}$,which implies $e^{-100bT} = \frac{1}{3}$.
After $200$ oscillations,the time elapsed is $t = 200T$.
The amplitude $a'$ will be $a' = a_0 e^{-b(200T)}$.
This can be written as $a' = a_0 (e^{-100bT})^2$.
Substituting the value from the first step: $a' = a_0 (\frac{1}{3})^2 = \frac{a_0}{9}$.
Thus,the amplitude becomes $\frac{1}{9}$ of the initial value.

(D) For a damped harmonic oscillator,the mechanical energy $E$ at time $t$ is given by $E(t) = E_0 e^{-bt/m}$,where $b$ is the damping constant and $m$ is the mass of the block.
We are given that the energy drops to half of its initial value,so $E(t) = E_0 / 2$.
Substituting this into the equation: $E_0 / 2 = E_0 e^{-bt/m}$.
This simplifies to $1/2 = e^{-bt/m}$,or $2 = e^{bt/m}$.
Taking the natural logarithm on both sides: $\ln(2) = bt/m$.
Solving for $t$: $t = (m/b) \ln(2)$.
Given $m = 0.1\, kg$ and $b = 10^{-2}\, kg\,s^{-1}$,we have $m/b = 0.1 / 10^{-2} = 10\, s$.
Thus,$t = 10 \times \ln(2) \approx 10 \times 0.693 = 6.93\, s$.
Rounding to the nearest provided option,the value is closest to $7\, s$.

(C) The equation of motion for a forced harmonic oscillator is given by $m \frac{d^2x}{dt^2} + m \omega^2 x = F(t)$.
Given $\omega = 2 \, rad \, s^{-1}$ and $F(t) = \sin(t)$,we have $\frac{d^2x}{dt^2} + 4x = \frac{1}{m} \sin(t)$.
Assuming $m = 1$ for proportionality,the particular solution is of the form $x_p = A \sin(t)$.
Substituting into the differential equation: $-A \sin(t) + 4A \sin(t) = \sin(t) \Rightarrow 3A = 1 \Rightarrow A = 1/3$.
However,considering the initial conditions $x(0) = 0$ and $v(0) = 0$,the general solution is $x(t) = c_1 \cos(2t) + c_2 \sin(2t) + \frac{1}{3} \sin(t)$.
Applying $x(0) = 0 \Rightarrow c_1 = 0$.
Applying $v(0) = 0 \Rightarrow v(t) = 2c_2 \cos(2t) + \frac{1}{3} \cos(t) \Rightarrow 2c_2 + 1/3 = 0 \Rightarrow c_2 = -1/6$.
Thus,$x(t) = \frac{1}{3} \sin(t) - \frac{1}{6} \sin(2t) = \frac{1}{3} (\sin(t) - \frac{1}{2} \sin(2t))$.
Therefore,the position is proportional to $\sin(t) - \frac{1}{2} \sin(2t)$.

(D) The energy of a damped oscillator at time $t$ is given by $E(t) = E_0 e^{-\gamma t}$,where $\gamma = \frac{b}{m}$ is the damping constant.
Given,initial energy $E_0 = 45\, J$ and final energy $E = 15\, J$.
The time taken for $15$ oscillations with a time period $T = 1\, s$ is $t = 15 \times T = 15 \times 1 = 15\, s$.
Substituting the values into the equation: $15 = 45 e^{-\gamma (15)}$.
Dividing by $45$: $\frac{1}{3} = e^{-15\gamma}$.
Taking the natural logarithm on both sides: $\ln(1/3) = -15\gamma$.
$-\ln 3 = -15\gamma$.
Therefore,$\gamma = \frac{1}{15} \ln 3\, s^{-1}$.

(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$,where $b$ is the damping constant. According to Stokes' law,$b = 6\pi \eta r$,where $\eta$ is the coefficient of viscosity.
For air: $8 = 10 e^{-(b_{air}/2m) \cdot 40} \implies 0.8 = e^{-(b_{air}/2m) \cdot 40}$.
Taking natural log: $\ln(0.8) = -(b_{air}/2m) \cdot 40 \implies \ln(4/5) = -(b_{air}/2m) \cdot 40 \implies \ln(5/4) = (b_{air}/2m) \cdot 40$.
So,$(b_{air}/2m) = \frac{\ln(1.25)}{40} = \frac{0.223}{40} = 0.005575 \ s^{-1}$.
Given $\frac{\eta_{air}}{\eta_{CO_2}} = 1.3$,then $b_{CO_2} = \frac{b_{air}}{1.3}$.
Thus,$(b_{CO_2}/2m) = \frac{b_{air}}{1.3 \cdot 2m} = \frac{0.005575}{1.3} \approx 0.004288 \ s^{-1}$.
For $CO_2$: $5 = 10 e^{-(b_{CO_2}/2m) \cdot t} \implies 0.5 = e^{-(0.004288)t}$.
Taking natural log: $\ln(0.5) = -0.004288 \cdot t \implies -0.693 = -0.004288 \cdot t$.
$t = \frac{0.693}{0.004288} \approx 161.6 \ s$.
Therefore,the time is close to $161 \ s$.

(A) The angular frequency of an undamped oscillator is $\omega_0 = \sqrt{\frac{k}{m}}$.
The angular frequency of a damped oscillator is $\omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}} = \omega_0 \sqrt{1 - \frac{r^2}{4mk}}$.
Using the binomial approximation $(1-x)^n \approx 1-nx$ for small $x$,we get $\omega \approx \omega_0 (1 - \frac{r^2}{8mk})$.
The time period $T = \frac{2\pi}{\omega}$,so $T \approx T_0 (1 - \frac{r^2}{8mk})^{-1} \approx T_0 (1 + \frac{r^2}{8mk})$.
The fractional change in time period is $\frac{\Delta T}{T_0} = \frac{T - T_0}{T_0} = \frac{r^2}{8mk}$.
Given $\frac{r^2}{mk} = 8\% = 0.08$,we have $\frac{\Delta T}{T_0} = \frac{0.08}{8} = 0.01 = 1\%$.
Since the value is positive,the time period increases by $1\%$.

(D) The amplitude of a damped harmonic oscillator is given by $A(t) = A_0 e^{-\gamma t}$.
Given frequency $f = 5 \text{ Hz}$,the time period of one oscillation is $T = \frac{1}{f} = 0.2 \text{ s}$.
For $10$ oscillations,the time taken is $t_{10} = 10 \times 0.2 = 2 \text{ s}$.
At $t = 2 \text{ s}$,the amplitude becomes half,so $A(2) = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-\gamma(2)} \implies 2 = e^{2\gamma} \implies \gamma = \frac{\ln 2}{2}$.
We need to find the time $t$ when $A(t) = \frac{A_0}{1000}$.
$\frac{A_0}{1000} = A_0 e^{-\gamma t} \implies 1000 = e^{\gamma t} \implies \ln(1000) = \gamma t$.
$3 \ln(10) = \left(\frac{\ln 2}{2}\right) t$.
$t = \frac{6 \ln(10)}{\ln 2} \approx \frac{6 \times 2.303}{0.693} \approx 19.94 \text{ s}$.
Thus,$t \approx 20 \text{ s}$.

(D) The displacement of a damped harmonic oscillator is given by $x(t) = A(t) \cos(\omega t + \varphi)$,where $A(t) = A_0 e^{-bt/2m}$.
Comparing this with the given equation $x(t) = e^{-0.1t} \cos(10\pi t + \varphi)$,we identify the amplitude as $A(t) = A_0 e^{-0.1t}$,where $A_0 = 1$.
We need to find the time $t$ when the amplitude $A(t)$ becomes half of its initial value $A_0$.
Set $A(t) = \frac{A_0}{2}$,which gives $A_0 e^{-0.1t} = \frac{A_0}{2}$.
Dividing both sides by $A_0$,we get $e^{-0.1t} = \frac{1}{2}$,or $e^{0.1t} = 2$.
Taking the natural logarithm on both sides: $0.1t = \ln(2)$.
Using the value $\ln(2) \approx 0.693$,we get $0.1t = 0.693$.
Therefore,$t = \frac{0.693}{0.1} = 6.93 \, s$.
Rounding to the nearest integer,we get $t \approx 7 \, s$.

(B) For a forced oscillator,the amplitude $A$ is given by the formula: $A = \frac{F_0}{\sqrt{m^2(\omega^2 - \omega_d^2)^2 + (b\omega_d)^2}}$,where $F_0$ is the driving force amplitude,$m$ is the mass,$\omega$ is the natural frequency,$\omega_d$ is the driving frequency,and $b$ is the damping constant.
When the driving frequency $\omega_d$ is close to the natural frequency $\omega$,the term $(\omega^2 - \omega_d^2)$ becomes very small,approaching zero.
In this condition,the amplitude is dominated by the damping term: $A \approx \frac{F_0}{\sqrt{(b\omega_d)^2}} = \frac{F_0}{b\omega_d}$.
Thus,the maximum possible amplitude is $\frac{F_0}{b\omega_d}$.

(A) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-\left(\frac{b}{2m}\right)t}$.
Given: $m = 200 \, g = 0.2 \, kg$,$b = 40 \, g \, s^{-1} = 0.04 \, kg \, s^{-1}$.
We want to find $t$ when $A(t) = \frac{A_0}{2}$.
Substituting the values: $\frac{A_0}{2} = A_0 e^{-\left(\frac{0.04}{2 \times 0.2}\right)t}$.
$\frac{1}{2} = e^{-\left(\frac{0.04}{0.4}\right)t} = e^{-0.1t}$.
Taking the natural logarithm on both sides: $\ln(0.5) = -0.1t$.
$-0.693 = -0.1t$.
$t = \frac{0.693}{0.1} = 6.93 \, s \approx 7 \, s$.

(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.
Given that at $t = 2 \, s$,the amplitude becomes $1/3$ of the original amplitude $A_0$:
$\frac{A_0}{3} = A_0 e^{-\frac{b(2)}{2m}}$
$\frac{1}{3} = e^{-\frac{b}{m}}$
Now,we need to find the amplitude at $t = 6 \, s$:
$A(6) = A_0 e^{-\frac{b(6)}{2m}} = A_0 (e^{-\frac{b}{m}})^3$
Substituting the value $e^{-\frac{b}{m}} = 1/3$:
$A(6) = A_0 \left(\frac{1}{3}\right)^3 = \frac{A_0}{27}$
Given that $A(6) = \frac{A_0}{n}$,we have:
$\frac{A_0}{n} = \frac{A_0}{27}$
$n = 27 = 3^3$.

(D) For a damped forced oscillator,the amplitude $A$ is given by $A = \frac{F_0}{\sqrt{m^2(\omega_0^2 - \omega^2)^2 + b^2\omega^2}}$.
The amplitude is maximum when the denominator is minimum,which occurs at $\omega_1 = \sqrt{\omega_0^2 - 2\gamma^2}$,where $\gamma = \frac{b}{2m}$.
The energy of the oscillator is proportional to the square of the amplitude $(E \propto A^2)$.
The energy is maximum when the amplitude is maximum,which occurs at $\omega_2 = \omega_0$ (the natural frequency) in the context of power absorption,or specifically,the resonance frequency for energy is $\omega_2 = \omega_0$.
Comparing the two,since $\omega_1 = \sqrt{\omega_0^2 - 2\gamma^2}$,it is clear that $\omega_1 < \omega_0$.
Therefore,$\omega_1 < \omega_2$.

(B) Resonance occurs when an external periodic force is applied to a system such that the frequency of the external force matches the natural frequency of the system.
Since resonance involves an external driving force causing the system to vibrate at a specific frequency,it is a special case of forced vibration.
Therefore,resonance is an example of forced vibration.

(C) The amplitude of an oscillating pendulum decreases with time due to air resistance (damping). This is known as damped oscillation.
However,the frequency of an oscillating pendulum depends on its length $L$ and acceleration due to gravity $g$,given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$.
Since $g$ and $L$ remain constant during the motion,the frequency of the pendulum remains constant over time.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) The $Assertion$ is correct. Resonance occurs when the driving frequency $(\omega)$ equals the natural frequency $(\omega_{0})$ of the system,resulting in maximum amplitude.
The $Reason$ is incorrect. The amplitude of forced vibrations does not simply increase with the frequency of the external force. Instead,the amplitude follows a resonance curve; it increases as the driving frequency approaches the natural frequency and decreases as it moves away from it.
The amplitude of oscillation for a forced,damped oscillator is given by:
$A = \frac{F_{0} / m}{\sqrt{(\omega^{2} - \omega_{0}^{2})^{2} + (b \omega / m)^{2}}}$
where $b$ is the damping constant and $\omega_{0} = \sqrt{k / m}$ is the natural frequency. As $\omega$ approaches $\omega_{0}$,the denominator decreases,causing the amplitude $A$ to reach its maximum value.

(N/A) Given: $m = 200 \; g = 0.2 \; kg$,$k = 90 \; N m^{-1}$,$b = 40 \; g s^{-1} = 0.04 \; kg s^{-1}$.
We check the condition for light damping: $\sqrt{km} = \sqrt{90 \times 0.2} = \sqrt{18} \approx 4.243 \; kg s^{-1}$.
Since $b \ll \sqrt{km}$,the period of oscillation $T$ is given by:
$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2}{90}} = 2\pi \sqrt{\frac{1}{450}} = \frac{2\pi}{21.21} \approx 0.296 \; s \approx 0.3 \; s$.
$(b)$ The amplitude $A(t)$ decays as $A(t) = A_0 e^{-bt/2m}$. For $A(t) = A_0/2$:
$\frac{1}{2} = e^{-bt/2m} \implies \ln(2) = \frac{bt}{2m} \implies t = \frac{2m \ln(2)}{b}$.
$t = \frac{2 \times 0.2 \times 0.693}{0.04} = \frac{0.2772}{0.04} = 6.93 \; s$.
$(c)$ The mechanical energy $E(t)$ decays as $E(t) = E_0 e^{-bt/m}$. For $E(t) = E_0/2$:
$\frac{1}{2} = e^{-bt/m} \implies \ln(2) = \frac{bt}{m} \implies t = \frac{m \ln(2)}{b}$.
$t = \frac{0.2 \times 0.693}{0.04} = \frac{0.1386}{0.04} = 3.465 \; s \approx 3.46 \; s$.

(A) Mass of the automobile,$m = 3000 \; kg$.
Displacement in the suspension system,$x = 15 \; cm = 0.15 \; m$.
There are $4$ springs in parallel to support the mass of the automobile.
The equation for the restoring force is $F = 4kx = mg$.
Thus,the spring constant $k$ for one wheel is $k = \frac{mg}{4x} = \frac{3000 \times 9.8}{4 \times 0.15} = 49000 \; N/m = 4.9 \times 10^4 \; N/m$.
For one wheel,mass $M = 750 \; kg$.
The time period of oscillation is $T = 2\pi \sqrt{\frac{M}{k}} = 2\pi \sqrt{\frac{750}{49000}} \approx 0.778 \; s$.
The amplitude of oscillation decreases by $50\%$,so $A = A_0 e^{-bt/2M} = 0.5 A_0$.
Taking natural log,$\ln(2) = \frac{bt}{2M}$.
$b = \frac{2M \ln(2)}{T} = \frac{2 \times 750 \times 0.693}{0.778} \approx 1336.5 \; kg/s$.

(N/A) Absolutely pure simple harmonic motion $(SHM)$ is not possible in reality because it requires the absence of all dissipative forces,such as friction,air resistance,and viscosity.
In any physical system,these dissipative forces are always present,which cause the energy of the oscillating system to dissipate over time.
This leads to damping,where the amplitude of the oscillation gradually decreases,eventually bringing the system to rest.
Therefore,$SHM$ is an idealization,while real-world systems exhibit damped harmonic motion.

(N/A) Pure simple harmonic oscillation is defined as the motion of a mechanical system where no resistive or internal frictional forces act on the system during its oscillation.
Such an oscillation is an ideal situation. In practice,it is impossible to achieve because any mechanical system oscillates within a medium (like air or liquid),which inevitably introduces resistive forces such as air drag or fluid viscosity. Additionally,internal friction within the components of the system always exists. These forces cause energy dissipation,leading to damping of the oscillations. Therefore,pure simple harmonic motion cannot be executed in reality.

(N/A) Damped oscillations: Oscillations in which the amplitude of the oscillator decreases with time are known as damped oscillations.
The motion of a simple pendulum swinging in air eventually dies out because air drag and friction at the support oppose the motion.
The mechanical energy of the oscillating system is dissipated as heat due to these resistive forces,causing the mechanical energy to decrease. According to the relation $E = \frac{1}{2} k A^2$,the amplitude $A$ gradually decreases.
Illustration with a spring:
Consider a block of mass $m$ connected to an elastic spring of spring constant $k$,oscillating vertically as shown in the figure. If the block is pushed down and released,it oscillates in a vertical plane with an angular frequency $\omega = \sqrt{\frac{k}{m}}$.
In practice,the surrounding medium (e.g.,air) exerts a damping force on the block,causing the mechanical energy of the block-spring system to decrease. This energy loss appears as heat in the surrounding medium and the block. The damping force depends on the nature of the medium; if the block is immersed in a liquid,the damping is much greater and energy dissipation is faster. The damping force is generally proportional to the velocity of the block.

(N/A) The resistive force acting on an oscillator in a fluid medium depends upon the velocity of the oscillator.
In practice,for not very large velocities,the resistive force is directly proportional to the velocity of the oscillator.
$\therefore F_{d} \propto -v$
$\therefore F_{d} = -bv$ ... $(1)$
Where $b$ is a positive constant called the damping coefficient.
It depends on the characteristics of the medium (like viscosity) and the size and shape of the block.
The unit of the damping coefficient is $N \cdot s/m$ and its dimensional formula is $[M^1 L^0 T^{-1}]$.
When the oscillator has a displacement $x$ from the mean position,the restoring force $(F_{s})$ is $F_{s} = -kx$ ... $(2)$
Thus,the total force acting on the mass at any time $t$ is:
$F = F_{s} + F_{d}$
$\therefore F = -kx - bv$ ... $(3)$
If $a(t)$ is the acceleration of the mass at time $t$,then by Newton's second law of motion,$F = ma(t)$.
Substituting into equation $(3)$:
$m \frac{d^2x}{dt^2} = -kx - b \frac{dx}{dt}$
$\therefore m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$ ... $(4)$
This is the differential equation for damped oscillations.
The solution of equation $(4)$ is:
$x(t) = A e^{-\frac{bt}{2m}} \cos(\omega' t + \phi)$ ... $(5)$
Where $A e^{-\frac{bt}{2m}}$ is the time-dependent amplitude and $\omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$ is the angular frequency of the damped oscillator.

(N/A) The mechanical energy of a simple harmonic oscillator $(SHO)$ is given by $E = \frac{1}{2} k A^{2}$.
For a damped oscillator,the amplitude at time $t$ is given by $A(t) = A e^{-\frac{b t}{2 m}}$.
Substituting this amplitude into the energy expression,the mechanical energy $E(t)$ of the damped oscillator at time $t$ is:
$E(t) = \frac{1}{2} k \left(A e^{-\frac{b t}{2 m}}\right)^{2} = \frac{1}{2} k A^{2} e^{-\frac{b t}{m}}$.
Thus,the mechanical energy is not constant but decreases exponentially with time.
This equation is valid for small damping where $b \ll \sqrt{k m}$,implying that the dimensionless ratio $\frac{b}{\sqrt{k m}} \ll 1$.
If $b = 0$,the expression reduces to the energy of an undamped oscillator,$E = \frac{1}{2} k A^{2}$.

(N/A) Damped oscillations are oscillations in which the amplitude of the oscillating system decreases over time due to the presence of dissipative forces,such as friction or air resistance.
In an ideal system,an oscillator would continue to vibrate indefinitely with constant amplitude. However,in real-world systems,energy is lost to the surroundings as heat or sound.
The equation of motion for a damped harmonic oscillator is given by $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$,where $m$ is the mass,$b$ is the damping constant,and $k$ is the spring constant.
As a result of these dissipative forces,the mechanical energy of the system gradually decreases,leading to the eventual cessation of motion.

(A) For an oscillator moving in a medium with a small velocity,the damping force $F_d$ is proportional to the velocity $v$ of the oscillator.
Mathematically,this is expressed as $F_d = -bv$,where $b$ is the damping constant.
Therefore,the damping force depends linearly on the velocity of the oscillator.

(N/A) For a damped harmonic oscillator,the equation of motion is given by $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$.
The amplitude of a damped oscillation decreases exponentially with time and is given by $A(t) = A_0 e^{-bt/2m}$,where $A_0$ is the initial amplitude,$b$ is the damping constant,$m$ is the mass,and $t$ is time.
The angular frequency of the damped oscillation is given by $\omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}$,where $k$ is the spring constant.

(A) For a damped harmonic oscillator,the angular frequency of oscillation is given by $\omega' = \sqrt{\omega_0^2 - (b/2m)^2}$,where $\omega_0$ is the natural frequency,$b$ is the damping constant,and $m$ is the mass.
When the damping is very heavy (overdamped),the term $(b/2m)^2$ becomes greater than or equal to $\omega_0^2$.
In the case of heavy damping,the system does not oscillate at all; it returns to the equilibrium position without crossing it.
Therefore,the frequency of oscillation becomes zero.

(A) For a damped harmonic oscillator,the amplitude $A$ at any time $t$ is given by the equation $A(t) = A_0 e^{-bt/2m}$,where $A_0$ is the initial amplitude,$b$ is the damping constant,$m$ is the mass of the oscillator,and $t$ is the time elapsed.
Given the time $t = 2m$,we substitute this value into the standard damping equation.
$A = A_0 e^{-b(2m)/2m}$
Simplifying the exponent,the $2m$ terms cancel out:
$A = A_0 e^{-b}$
However,looking at the provided options,the question asks for the general form where $t$ is replaced by $2m$ in the exponent. The standard formula is $A = A_0 e^{-bt/2m}$. Substituting $t = 2m$ into the expression $bt/2m$ yields $b(2m)/2m = b$. Thus,the amplitude is $A_0 e^{-b}$. Since the options represent the functional form,option $A$ is the correct representation of the damped amplitude formula.

(N/A) Natural Oscillation: When an oscillator is slightly disturbed from its equilibrium position and released,it starts oscillating. The oscillations performed by it in the absence of any type of resistive force are known as natural oscillations.
The angular frequency of natural oscillation is $\omega_{0}$ and the frequency is $f_{0}$.
Free Oscillation: When a system is displaced from its equilibrium position and released,it oscillates with its natural frequency. These oscillations are known as free oscillations.
The angular frequency of free oscillation is $\omega$.
Forced (driven) Oscillations: The oscillations executed by a system under the influence of an external periodic force are called forced oscillations.
The angular frequency of forced oscillation is called $\omega_{d}$,which is the angular frequency of the external force.
In practice,oscillations always occur in some medium; hence,some type of damping force is always acting on the system,and eventually,the oscillations die out with time. To sustain the oscillations,an external periodic force is required.
Example: When a child on a garden swing periodically presses their feet against the ground.

(N/A) To sustain oscillations in a system,an external periodic force is applied,given by:
$F(t) = F_{0} \cos \omega_{d} t$
where $F_{0}$ is the amplitude and $\omega_{d}$ is the angular frequency of the driving force.
Three forces act on the oscillator:
$(1)$ Restoring force: $F_{r} = -k x(t)$
$(2)$ Resistive (damping) force: $F_{s} = -b v(t) = -b \frac{dx}{dt}$
$(3)$ External periodic force: $F_{d} = F_{0} \cos \omega_{d} t$
According to Newton's second law of motion,the net force is $F_{net} = ma(t) = m \frac{d^{2}x}{dt^{2}}$.
Thus,$m \frac{d^{2}x}{dt^{2}} = F_{r} + F_{s} + F_{d}$
$m \frac{d^{2}x}{dt^{2}} = -kx - b \frac{dx}{dt} + F_{0} \cos \omega_{d} t$
Rearranging the terms,we get the differential equation of forced oscillation:
$m \frac{d^{2}x}{dt^{2}} + b \frac{dx}{dt} + kx = F_{0} \cos \omega_{d} t$
Dividing by mass $m$,we obtain:
$\frac{d^{2}x}{dt^{2}} + \frac{b}{m} \frac{dx}{dt} + \frac{k}{m} x = \frac{F_{0}}{m} \cos \omega_{d} t$

(N/A) For small damping,the amplitude $A$ of a forced oscillator is given by $A = \frac{F_0}{\sqrt{m^2(\omega^2 - \omega_d^2)^2 + (\omega_d b)^2}}$.
When the driving frequency $\omega_d$ is far from the natural frequency $\omega$,such that $\omega_d b << m|\omega^2 - \omega_d^2|$,the term $(\omega_d b)^2$ can be neglected in the denominator.
Thus,the amplitude simplifies to $A \approx \frac{F_0}{m|\omega^2 - \omega_d^2|}$.
In this regime,the amplitude is primarily determined by the stiffness and inertia of the system rather than the damping constant $b$. As $\omega_d$ moves further away from $\omega$,the amplitude decreases significantly.
When $\omega_d = \omega$,the amplitude is limited only by the damping term,$A = \frac{F_0}{\omega_d b}$. If $b = 0$,the amplitude becomes infinite at resonance. As damping $b$ increases,the peak amplitude decreases and shifts slightly.

(N/A) The amplitude $A$ of a forced oscillator is given by the expression:
$A = \frac{F_{0}}{\left[m^{2}(\omega_{0}^{2} - \omega^{2})^{2} + (\omega b)^{2}\right]^{1/2}}$
where $F_{0}$ is the driving force amplitude,$m$ is the mass,$\omega_{0}$ is the natural frequency,$\omega$ is the driving frequency,and $b$ is the damping constant.
When the driving frequency $\omega$ is very close to the natural frequency $\omega_{0}$,the term $m^{2}(\omega_{0}^{2} - \omega^{2})^{2}$ becomes very small compared to $(\omega b)^{2}$.
In this condition,the amplitude is approximately $A \approx \frac{F_{0}}{\omega b}$.
Since the damping constant $b$ is non-zero,the amplitude remains finite and does not reach infinity.
Resonance: The phenomenon where the amplitude of oscillation increases significantly when the driving frequency of an external force is close to the natural frequency of the oscillator is called resonance.

(N/A) Consider a set of five simple pendulums of assorted lengths suspended from a common horizontal rope,as shown in the figure.
Pendulums-$1$ and $4$ have the same lengths,while the others have different lengths. Let us set pendulum-$1$ into motion. The energy from this pendulum is transferred to the other pendulums through the connecting rope,causing them to start oscillating.
Pendulums-$2$,$3$,and $5$ initially start oscillating with their own natural frequencies and different amplitudes,but this motion is gradually damped and not sustained. Their frequencies of oscillation gradually change,and ultimately,they oscillate with the frequency of pendulum-$1$ (the driving frequency),but with different amplitudes.
Pendulum-$4$ behaves differently from the others. Since it has the same length as pendulum-$1$,it has the same natural frequency. It oscillates with the same frequency as pendulum-$1$,and its amplitude gradually increases,becoming very large. This is the phenomenon of resonance.
In general,a system may have several natural frequencies,for example,vibrating strings,air columns,etc.

(N/A) Resonance occurs when the frequency of an external periodic force matches the natural frequency of a system,leading to a significant increase in the amplitude of oscillations.
Examples of Resonance:
$(1)$ Soldiers marching on a suspended bridge are advised to break their step. If the frequency of their marching matches the natural frequency of the bridge,resonance occurs,causing the amplitude of the bridge's vibrations to increase drastically,which can lead to the collapse of the bridge.
$(2)$ During the design of bridges,engineers ensure that the natural frequency of the bridge does not match the frequency of external forces caused by wind gusts. If these frequencies coincide,the resulting resonance can cause structural failure.
$(3)$ During an earthquake,buildings of different heights respond differently to seismic waves. Medium-height structures are often more susceptible to damage because their natural frequencies are closer to the frequencies of seismic waves,leading to resonance,whereas low-rise and high-rise structures have natural frequencies that are significantly different from the seismic wave frequencies.