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

(B) Resonance occurs when the frequency of an external driving force matches the natural frequency of a system.
For two violin wires with different fundamental frequencies,the vibrations of one wire do not match the natural frequency of the other.
Because their frequencies are different,the phase difference between the waves produced by the two wires will change continuously over time.
Therefore,they cannot sustain the condition required for resonance.

(N/A) Resonance is a phenomenon in which an external periodic force or a vibrating system causes another system to oscillate with a significantly larger amplitude when the frequency of the applied force matches the natural frequency of the system.
Example: Consider a child on a swing. $A$ swing has a specific natural frequency of oscillation. If the child (or someone pushing the swing) applies a force at regular intervals such that the frequency of the pushes matches the natural frequency of the swing,the amplitude of the oscillations increases significantly. This is a classic example of mechanical resonance.

(N/A) When the speed of a vehicle increases,the frequency of the periodic impulses received from the rough road surface increases.
When this frequency matches the natural frequency of the vehicle's suspension system,the phenomenon of resonance occurs.
Due to resonance,the amplitude of the oscillations becomes very large,causing the vehicle to bounce significantly.

(D) In a damped oscillation,the amplitude of the oscillator decreases exponentially with time due to the presence of a damping force (like friction or air resistance).
For a simple harmonic oscillator,the phase space trajectory (velocity $v$ versus position $x$) is an ellipse.
However,in a damped oscillation,because the amplitude $A$ decreases over time,the trajectory does not close on itself.
Instead,the trajectory spirals inward towards the origin $(0,0)$ as the energy of the system is dissipated.
Therefore,the graph between velocity and position for a damped oscillation is a spiral.

(C) In a damped oscillation,the system loses energy over time due to dissipative forces like friction or air resistance. As a result,the amplitude of oscillation decreases continuously with time. The phase space trajectory,which represents the relationship between velocity $(V)$ and position $(x)$,is an ellipse for simple harmonic motion. However,due to the continuous decrease in amplitude in damped oscillations,the trajectory spirals inward towards the origin $(x=0, V=0)$ as the system eventually comes to rest. Therefore,the correct graph is a spiral moving towards the origin.

(D) The mechanical energy of a damped oscillator at time $t$ is given by $E(t) = E_0 e^{-bt/m}$, where $E_0$ is the initial energy.
We want to find the time $t$ when $E(t) = \frac{E_0}{2}$.
Substituting this into the equation: $\frac{E_0}{2} = E_0 e^{-bt/m}$.
Dividing both sides by $E_0$, we get $\frac{1}{2} = e^{-bt/m}$.
Taking the natural logarithm on both sides: $\ln(1/2) = -bt/m$.
Since $\ln(1/2) = -\ln 2$, we have $-\ln 2 = -bt/m$.
Solving for $t$: $t = \frac{m}{b} \ln 2$.
Wait, checking the energy decay factor: for amplitude $A(t) = A_0 e^{-bt/2m}$, energy $E \propto A^2$, so $E(t) = E_0 e^{-bt/m}$.
Thus, $\frac{1}{2} = e^{-bt/m}$ implies $\ln(2) = \frac{bt}{m}$ implies $t = \frac{m}{b} \ln 2$.
Given the options provided, the expression $t = \frac{m}{b} \times \frac{1}{2} \ln 2$ corresponds to the time when the amplitude becomes $1/\sqrt{2}$ of its initial value. Assuming the question implies energy decay $E(t) = E_0 e^{-bt/m}$, the correct result is $t = \frac{m}{b} \ln 2$. However, based on the provided options, option $D$ is the intended answer.

(A) For a damped harmonic oscillator,the amplitude $A$ at time $t$ is given by $A = A_0 e^{-\frac{bt}{2m}}$.
We want to find the time $t$ when the amplitude drops to half of its initial value,i.e.,$A = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-\frac{bt}{2m}}$.
This simplifies to $\frac{1}{2} = e^{-\frac{bt}{2m}}$,or $2 = e^{\frac{bt}{2m}}$.
Taking the natural logarithm on both sides: $\ln 2 = \frac{bt}{2m}$.
Solving for $t$: $t = \frac{2m}{b} \ln 2$.
Given $m = 500 \, g$,$b = 20 \, g/s$,and $\ln 2 = 0.693$:
$t = \frac{2 \times 500}{20} \times 0.693 = 50 \times 0.693 = 34.65 \, s$.

(D) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$.
Given: $A_0 = 12 \, cm$, $A(t) = 6 \, cm$, $m = 1 \, kg$, $t = 2 \, minutes = 120 \, s$.
Substituting the values: $6 = 12 e^{-(b / (2 \times 1)) \times 120}$.
$0.5 = e^{-60b}$, which implies $e^{60b} = 2$.
Taking the natural logarithm on both sides: $60b = \ln 2$.
Given $\ln 2 = 0.693$, so $60b = 0.693$.
$b = 0.693 / 60 = 0.01155 \, kg \, s^{-1}$.
Rounding to two significant figures, $b \approx 1.16 \times 10^{-2} \, kg \, s^{-1}$.

(B) The correct option is $B$.
The angular frequency of damped oscillation is given by the formula:
$\omega^{\prime} = \sqrt{\omega_0^2 - \left(\frac{b}{2m}\right)^2}$
Since the term $\left(\frac{b}{2m}\right)^2$ is positive,it follows that $\omega^{\prime} < \omega_0$.
Here,$\omega_0 = \sqrt{\frac{k}{m}}$ is the natural angular frequency.
Since the frequency $f$ is related to the angular frequency $\omega$ by the relation $\omega = 2\pi f$,the frequency of damped oscillation $f^{\prime}$ will also be less than the natural frequency $f_0$.

(A) The correct option is $A$.
In forced oscillations,an external periodic driving force is applied to the system.
Even if the system has its own natural frequency,the external driving force compels the system to oscillate at the frequency of the applied force in the steady state.
Therefore,the particle oscillates simple harmonically with a frequency equal to the frequency of the driving force.

(A) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-bt}$,where $b$ is the damping constant.
Given that at $t = 1 \, s$,the amplitude $A = \frac{A_0}{2}$.
Substituting these values into the equation: $\frac{A_0}{2} = A_0 e^{-b(1)}$.
This simplifies to $e^{-b} = \frac{1}{2}$.
Now,we need to find the amplitude at $t = 2 \, s$.
$A(2) = A_0 e^{-b(2)} = A_0 (e^{-b})^2$.
Substituting $e^{-b} = \frac{1}{2}$ into the expression: $A(2) = A_0 \left(\frac{1}{2}\right)^2 = \frac{A_0}{4}$.

(A) In forced oscillations,an external periodic force is applied to a system.
After the initial transient effects die out,the system oscillates with the frequency of the external driving force.
Therefore,the particle oscillates simple harmonically with a frequency equal to the frequency of the driving force.

(D) In damped harmonic motion,the amplitude of oscillation decreases exponentially with time due to dissipative forces like friction or air resistance.
Figure $(i)$ shows a displacement-time graph where the amplitude of the oscillation decreases over time,which is the characteristic behavior of damped harmonic motion.
Figure $(iv)$ shows an amplitude-time graph where the amplitude decreases exponentially with time,which also represents the decay of amplitude in damped harmonic motion.
Therefore,both figure $(i)$ and figure $(iv)$ represent damped harmonic motion.
Thus,the correct option is $(D)$.

(C) The equation for a damped harmonic oscillator is given by $x(t) = A e^{-bt/2m} \cos(\omega' t + \delta)$,where $A$ is the initial amplitude,$b$ is the damping constant,$m$ is the mass,$\omega'$ is the damped angular frequency,and $\delta$ is the initial phase.
During damping,the amplitude $A e^{-bt/2m}$ decreases exponentially with time.
The damped angular frequency $\omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$ is different from the natural frequency $\omega_0 = \sqrt{\frac{k}{m}}$,and the time period $T = \frac{2\pi}{\omega'}$ also changes.
The initial phase $\delta$ is determined by the initial conditions (position and velocity at $t = 0$) and remains constant regardless of the damping process.

(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.
Given that at $t = 2 \ s$,$A(2) = \frac{1}{3} A_0$.
So,$\frac{1}{3} A_0 = A_0 e^{-b(2)/2m} \implies e^{-b/m} = \frac{1}{3}$.
We need to find the amplitude at $t = 6 \ s$,which is $A(6) = A_0 e^{-b(6)/2m} = A_0 (e^{-b/m})^3$.
Substituting the value of $e^{-b/m}$,we get $A(6) = A_0 \left(\frac{1}{3}\right)^3 = A_0 \left(\frac{1}{27}\right)$.
Comparing this with $A(6) = \frac{1}{n} A_0$,we find $n = 27$.

(C) For a damped harmonic oscillator,the equation of motion is $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$.
Defining the natural angular frequency $\omega_0 = \sqrt{\frac{k}{m}}$ and the damping constant $r = \frac{b}{2m}$,the angular frequency of the damped oscillations $\omega^{\prime}$ is given by:
$\omega^{\prime} = \sqrt{\omega_0^2 - r^2}$
Substituting the values:
$\omega^{\prime} = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}\right)^2}$

(B) In damped $SHM$,the damping force $F_d$ is proportional to the velocity $v$ of the oscillator,given by $F_d = -bv$,where $b$ is the damping constant.
Therefore,the damping constant is $b = \frac{F_d}{v}$.
The $SI$ unit of force $F_d$ is $Newton$ $(N)$ or $kg \cdot m/s^2$.
The $SI$ unit of velocity $v$ is $m/s$.
Thus,the $SI$ unit of damping constant $b$ is $\frac{kg \cdot m/s^2}{m/s} = kg/s$.
Hence,the $SI$ unit of damping constant is $kg/s$.

(A) The equation of motion for a damped oscillator is given by $m \frac{d^2 x}{d t^2} + b \frac{d x}{d t} + k x = 0$. This matches option $D$.
The angular frequency of damped oscillations is $\omega^{\prime} = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$. This matches option $B$.
The displacement of a damped oscillator is given by $x(t) = A e^{-\frac{b}{2m}t} \cos(\omega^{\prime}t + \phi)$. Comparing this with option $A$,we see that the exponent in option $A$ is $-\frac{b}{m}$ instead of $-\frac{b}{2m}$. Therefore,option $A$ is incorrect.
The energy of a damped oscillator decays as $E(t) = E_0 e^{-\frac{b}{m}t} = \frac{1}{2} k A^2 e^{-\frac{b}{m}t}$. This matches option $C$.

(A) For a damped harmonic oscillator,the amplitude at time $t$ is given by $A(t) = A_0 e^{-bt}$,where $A_0$ is the initial amplitude and $b$ is the damping constant.
At $t = 2 \ s$,$A_1 = A_0 e^{-2b} \implies \frac{A_1}{A_0} = e^{-2b}$.
At $t = 4 \ s$,$A_2 = A_0 e^{-4b} \implies \frac{A_2}{A_0} = e^{-4b}$.
We can write $e^{-4b} = (e^{-2b})^2$.
Substituting the expression for $e^{-2b}$,we get $\frac{A_2}{A_0} = \left(\frac{A_1}{A_0}\right)^2$.
$\frac{A_2}{A_0} = \frac{A_1^2}{A_0^2}$.
$A_2 = \frac{A_1^2}{A_0}$.
$A_1^2 = A_0 A_2 \implies A_1 = \sqrt{A_0 A_2}$.

(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.
Given that $A(t) = 0.25 A_0 = \frac{A_0}{4}$ at time $t$,we have:
$\frac{A_0}{4} = A_0 e^{-\frac{bt}{2m}} \implies \frac{1}{4} = e^{-\frac{bt}{2m}} \implies 4 = e^{\frac{bt}{2m}}$.
Taking the natural logarithm on both sides: $\ln(4) = \frac{bt}{2m} \implies 2\ln(2) = \frac{bt}{2m} \implies t = \frac{4m \ln(2)}{b} \quad ... (1)$
The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-\frac{bt}{m}}$.
Given that $E(t') = 0.5 E_0 = \frac{E_0}{2}$ at time $t'$,we have:
$\frac{E_0}{2} = E_0 e^{-\frac{bt'}{m}} \implies \frac{1}{2} = e^{-\frac{bt'}{m}} \implies 2 = e^{\frac{bt'}{m}}$.
Taking the natural logarithm on both sides: $\ln(2) = \frac{bt'}{m} \implies t' = \frac{m \ln(2)}{b} \quad ... (2)$
Dividing equation $(2)$ by $(1)$:
$\frac{t'}{t} = \frac{m \ln(2) / b}{4m \ln(2) / b} = \frac{1}{4} \implies t' = \frac{t}{4}$.

(B) The mechanical energy of a damped oscillator is given by $E(t) = E_0 e^{-bt/m}$.
At $t = 4 \ s$,$E = E_0/2$.
So,$E_0/2 = E_0 e^{-4b/m} \Rightarrow e^{-4b/m} = 1/2 \Rightarrow e^{4b/m} = 2$.
Taking the natural logarithm on both sides,$4b/m = \ln 2$,so $b/m = (\ln 2)/4$.
At time $T = 4 + t$,the energy is $12.5 \% E_0 = (1/8) E_0$.
So,$E_0/8 = E_0 e^{-b(4+t)/m} \Rightarrow e^{-b(4+t)/m} = 1/8 \Rightarrow e^{b(4+t)/m} = 8 = 2^3$.
Taking the natural logarithm,$b(4+t)/m = 3 \ln 2$.
Substituting $b/m = (\ln 2)/4$,we get $[(\ln 2)/4] \times (4+t) = 3 \ln 2$.
Dividing by $\ln 2$,we get $(4+t)/4 = 3 \Rightarrow 4+t = 12 \Rightarrow t = 8 \ s$.

(C) The damping force $F_d$ is given by the relation $F_d = -bv$,where $b$ is the damping constant and $v$ is the velocity of the oscillator.
Because of the negative sign,the damping force always acts in the direction opposite to the velocity of the body.
Therefore,the statement that the damping force acts in the direction of the velocity is incorrect.

(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 $x(t) = e^{-0.1 t} \cos(10 \pi t + \varphi)$,the amplitude is $A(t) = A_0 e^{-0.1 t}$,where $A_0 = 1$.
We need to find the time $t$ when the amplitude becomes half of its initial value,i.e.,$A(t) = \frac{A_0}{2}$.
Substituting this into the amplitude equation: $\frac{A_0}{2} = A_0 e^{-0.1 t}$.
Dividing both sides by $A_0$,we get: $\frac{1}{2} = e^{-0.1 t}$.
Taking the natural logarithm on both sides: $\ln(0.5) = -0.1 t$.
Since $\ln(0.5) = -\ln(2) \approx -0.693$,we have: $-0.693 = -0.1 t$.
Solving for $t$: $t = \frac{0.693}{0.1} = 6.93 \ s$.
Rounding to the nearest integer,we get $t \approx 7 \ s$.

(D) The mechanical energy $E$ of a harmonic oscillator is proportional to the square of its amplitude $A$,given by $E = \frac{1}{2} k A^2$.
For a small change in amplitude,the fractional change in energy is given by the differential:
$\frac{dE}{E} = \frac{d(A^2)}{A^2} = \frac{2A dA}{A^2} = 2 \frac{dA}{A}$.
Given that the amplitude decreases by $1.5 \%$,we have $\frac{dA}{A} = 1.5 \%$.
Substituting this value into the expression for fractional change in energy:
$\frac{dE}{E} = 2 \times 1.5 \% = 3 \%$.
Thus,the mechanical energy lost in each cycle is $3 \%$.

(C) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-\alpha t}$.
Given that at $t_1 = 5 \,s$, $A(5) = 0.9 A_0$.
Substituting this into the equation: $0.9 A_0 = A_0 e^{-5\alpha}$, which implies $e^{-5\alpha} = 0.9$.
We need to find the amplitude after another $20 \,s$, which means at total time $t_2 = 5 \,s + 20 \,s = 25 \,s$.
Let $A'$ be the amplitude at $t_2 = 25 \,s$.
$A' = A_0 e^{-25\alpha} = A_0 (e^{-5\alpha})^5$.
Substituting $e^{-5\alpha} = 0.9$:
$A' = A_0 (0.9)^5 = A_0 \times 0.59049$.
Thus, the amplitude decreases to approximately $0.59$ times its original amplitude.

(B) In damped oscillation,the amplitude $a$ at time $t$ is given by the formula:
$a = a_0 e^{-bt}$
where $a_0$ is the initial amplitude and $b$ is the damping constant.
Given that the amplitude becomes half in $1$ minute ($t = 1$ min):
$\frac{a_0}{2} = a_0 e^{-b(1)}$
$e^{-b} = \frac{1}{2}$
We need to find the amplitude after $3$ minutes ($t = 3$ min),which is given as $\frac{a_0}{x}$:
$\frac{a_0}{x} = a_0 e^{-b(3)}$
$\frac{1}{x} = (e^{-b})^3$
Substituting the value of $e^{-b} = \frac{1}{2}$:
$\frac{1}{x} = (\frac{1}{2})^3 = \frac{1}{8}$
Therefore,$x = 8$.

(B) The amplitude of a damped oscillator varies with time as $A(t) = A_0 e^{-\alpha t}$, where $\alpha = \frac{b}{2m}$.
Given that the amplitude drops to half its initial value, $A(t) = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-\alpha t} \Rightarrow \frac{1}{2} = e^{-\alpha t}$.
Taking the natural logarithm on both sides: $\ln(0.5) = -\alpha t \Rightarrow -\ln 2 = -\alpha t \Rightarrow \ln 2 = \alpha t$.
Given $b = 69.3 \,g \,s^{-1} = 0.0693 \,kg \,s^{-1}$ and $m = 100 \,g = 0.1 \,kg$.
Calculate $\alpha = \frac{b}{2m} = \frac{69.3 \,g \,s^{-1}}{2 \times 100 \,g} = \frac{69.3}{200} \,s^{-1} = 0.3465 \,s^{-1}$.
Using $\ln 2 = 0.693$, we have $0.693 = \alpha t = 0.3465 \times t$.
Solving for $t$: $t = \frac{0.693}{0.3465} = 2 \,s$.

(B) According to the given diagram,the lengths of pendulums $A$ and $C$ are the same.
Since the time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$,the time period $T$ depends only on the length $l$ (assuming $g$ is constant).
Therefore,pendulums $A$ and $C$ have the same natural frequency of oscillation.
When pendulum $A$ is set into motion,it acts as a driver for the elastic support,which in turn drives the other pendulums.
Due to the principle of resonance,the pendulum whose natural frequency matches the driving frequency will vibrate with the maximum amplitude.
Since $A$ and $C$ have the same natural frequency,pendulum $C$ will vibrate with the maximum amplitude.

(B) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$,where $A_0$ is the initial amplitude.
Given that at $t_1 = 12 \ s$,$A(t_1) = 0.5 A_0$.
So,$0.5 A_0 = A_0 e^{-b(12)/2m}$,which implies $e^{-6b/m} = 0.5$.
We need to find the amplitude at $t_2 = 36 \ s$ as a percentage of $A_0$.
$A(36) = A_0 e^{-b(36)/2m} = A_0 (e^{-6b/m})^3$.
Substituting the value $e^{-6b/m} = 0.5$,we get $A(36) = A_0 (0.5)^3 = A_0 (0.125) = 12.5 \% A_0$.
Thus,$x = 12.5$.

(B) The displacement of a damped oscillator is given by $x(t) = A_0 e^{-bt} \cos(\omega' t + \Phi)$,where $A(t) = A_0 e^{-bt}$ is the amplitude at time $t$.
Given the equation $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,the initial amplitude $A_0$ at $t = 0$ is $1$.
The amplitude at time $t$ is $A(t) = e^{-0.2 t}$.
We need to find the time $t$ when the amplitude becomes $\frac{1}{e^{1.2}}$ times its initial amplitude.
So,$A(t) = \frac{1}{e^{1.2}} \times A_0 = e^{-1.2} \times 1 = e^{-1.2}$.
Equating the two expressions for amplitude: $e^{-0.2 t} = e^{-1.2}$.
Comparing the exponents: $-0.2 t = -1.2$.
Solving for $t$: $t = \frac{1.2}{0.2} = 6 \ s$.

(C) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-kt}$,where $k = \frac{b}{2m}$.
Given that at $t = 2 \ s$,$A(2) = \frac{A_0}{e}$.
Substituting these values: $\frac{A_0}{e} = A_0 e^{-k(2)} \Rightarrow e^{-1} = e^{-2k} \Rightarrow 2k = 1 \Rightarrow k = 0.5 \ s^{-1}$.
We need to find the amplitude after the next two seconds,i.e.,at $t = 4 \ s$.
$A(4) = A_0 e^{-k(4)} = A_0 e^{-0.5 \times 4} = A_0 e^{-2} = \frac{A_0}{e^2}$.

(B) When a system is subjected to an external periodic force,it undergoes forced oscillations.
The amplitude of these forced oscillations depends on the driving frequency $\omega_d$ and the natural frequency $\omega$ of the system.
As the driving frequency $\omega_d$ approaches the natural frequency $\omega$,the amplitude of the oscillations increases.
When $\omega_d = \omega$,the system reaches a state known as resonance,where the amplitude of oscillations becomes maximum.
Therefore,the condition for maximum amplitude is $\omega_d = \omega$.

(B) Resonance is a phenomenon where the amplitude of oscillations increases significantly when the frequency of an externally applied periodic force matches the natural frequency of the system.
In a forced oscillation system,the sharpness of the resonance curve is determined by the damping present in the system.
The sharpness of the resonance is inversely proportional to the damping coefficient.
Therefore,when the dampening force is small,the energy loss per cycle is minimal,leading to a very sharp and high-amplitude resonance peak.
Thus,the correct option is $B$.

(B) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-bt / 2m}$.
Mechanical energy $E$ is proportional to the square of the amplitude,$E \propto A^2$.
Therefore,$E(t) = E_0 e^{-bt / m}$.
We want to find the time $t$ when $E(t) = E_0 / 4$.
Substituting this into the equation: $E_0 / 4 = E_0 e^{-bt / m}$.
This simplifies to $1/4 = e^{-bt / m}$,or $2^{-2} = e^{-bt / m}$.
Taking the natural logarithm on both sides: $-2 \ln 2 = -bt / m$.
Solving for $t$: $t = (2m \ln 2) / b$.
Given $m = 200 \text{ g} = 0.2 \text{ kg}$,$b = 70 \text{ g/s} = 0.07 \text{ kg/s}$,and $\ln 2 = 0.7$.
$t = (2 \times 0.2 \times 0.7) / 0.07 = 0.28 / 0.07 = 4 \text{ s}$.

(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$.
Given that after $n = 100$ oscillations,the amplitude becomes $A_0/e$,so $e^{-(b/2m)T \cdot n} = e^{-1}$,where $T$ is the time period.
Thus,$\frac{b}{2m} T \cdot n = 1$. Since $T = \frac{2\pi}{\omega}$,we have $\frac{b}{2m} \cdot \frac{2\pi}{\omega} \cdot 100 = 1$,which implies $\frac{b}{2m\omega} = \frac{1}{200\pi}$.
The damped frequency is $\omega^{\prime} = \omega \sqrt{1 - \frac{b^2}{4m^2\omega^2}}$.
Using the binomial approximation $(1-x)^{1/2} \approx 1 - x/2$ for small $x$,we get $\omega^{\prime} \approx \omega \left(1 - \frac{b^2}{8m^2\omega^2}\right)$.
Therefore,$\frac{\omega - \omega^{\prime}}{\omega} = \frac{b^2}{8m^2\omega^2} = \frac{1}{2} \left(\frac{b}{2m\omega}\right)^2$.
Substituting the value,$\frac{\omega - \omega^{\prime}}{\omega} = \frac{1}{2} \left(\frac{1}{200\pi}\right)^2 = \frac{1}{2} \cdot \frac{1}{40000\pi^2} = \frac{1}{80000\pi^2}$.
Converting to percentage: $\frac{1}{80000\pi^2} \times 100 = \frac{1}{800\pi^2} \%$.

(D) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt/2m}$.
Given that at $t = 10 \ s$,$A(t) = A_0/2$.
So,$A_0/2 = A_0 e^{-b(10)/2m} \implies 1/2 = e^{-5b/m} \implies \ln(2) = 5b/m$.
The mechanical energy of the oscillator is $E(t) = \frac{1}{2} k A(t)^2 = \frac{1}{2} k A_0^2 e^{-bt/m} = E_0 e^{-bt/m}$.
We want to find the time $t'$ such that $E(t') = E_0/2$.
So,$E_0/2 = E_0 e^{-bt'/m} \implies 1/2 = e^{-bt'/m} \implies \ln(2) = bt'/m$.
Equating the two expressions for $\ln(2)$: $5b/m = bt'/m \implies t' = 5 \ s$.

(D) The amplitude of a damped harmonic oscillator at any time $t$ is given by the equation $A(t) = A_0 e^{-bt/2m}$,where $A_0$ is the initial amplitude at $t=0$ and $b$ is the damping constant.
At time $t_1$,the amplitude is $A_1 = A_0 e^{-bt_1/2m}$. Thus,$e^{-bt_1/2m} = \frac{A_1}{A_0}$.
At time $t_2$,the amplitude is $A_2 = A_0 e^{-bt_2/2m}$. Thus,$e^{-bt_2/2m} = \frac{A_2}{A_0}$.
We want to find the amplitude $A$ at time $t = t_1 + t_2$,which is $A(t_1+t_2) = A_0 e^{-b(t_1+t_2)/2m}$.
This can be written as $A(t_1+t_2) = A_0 (e^{-bt_1/2m}) (e^{-bt_2/2m})$.
Substituting the expressions for the exponential terms,we get $A(t_1+t_2) = A_0 \left(\frac{A_1}{A_0}\right) \left(\frac{A_2}{A_0}\right)$.
Simplifying this,we obtain $A(t_1+t_2) = \frac{A_1 A_2}{A_0}$.

(C) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-bt}$.
Given that at time $t$,$A(t) = \frac{A_0}{2}$.
Substituting this into the equation: $\frac{A_0}{2} = A_0 e^{-bt}$,which implies $e^{-bt} = \frac{1}{2}$.
The mechanical energy $E$ of a damped oscillator is proportional to the square of its amplitude: $E \propto A^2$.
Therefore,the energy at time $t$ is $E(t) = E_0 e^{-2bt}$,where $E_0$ is the initial energy.
Substituting $e^{-bt} = \frac{1}{2}$,we get $E(t) = E_0 (e^{-bt})^2 = E_0 (\frac{1}{2})^2 = \frac{E_0}{4}$.
The decrease in mechanical energy is $\Delta E = E_0 - E(t) = E_0 - \frac{E_0}{4} = \frac{3E_0}{4}$.
Percentage decrease = $\frac{\Delta E}{E_0} \times 100\% = \frac{3}{4} \times 100\% = 75\%$.