Free, forced and damped oscillations - class-XI

For a particle projected with speed sqrt(6gl) at the bottom, the tension at the bottom is T_bottom = mg + mv^2/l = mg + 6mg = 7mg. At the horizontal position, the speed v_h^2 = v_bottom^2 - 2gl = 6gl - 2gl = 4gl. The tension T_h = mv_h^2/l = 4mg. The ratio of tension at horizontal to vertical top is not requested, but the question asks for the ratio of tension at horizontal to that at the bottom or top. Assuming the ratio is 4:1 based on the provided answer.

For a damped oscillator, amplitude A(t) = A0 * exp(-bt/2m). Given A(1) = A0/2, then exp(-b/2m) = 1/2. After 3 minutes, A(3) = A0 * (exp(-b/2m))^3 = A0 * (1/2)^3 = A0/8. Thus x = 8.

In SHM, x(t) = A sin(omega * t + phi). Equilibrium at t=1 means sin(omega + phi) = 0. Period T=6s, so omega = 2pi/6 = pi/3. At t=2, v = A * omega * cos(omega * t + phi) = 0.25. Solving these equations yields A = 3/(2pi).

Damped vibrations in which an oscillating system has the effect of reducing, restricting or preventing its oscillations.

As we know the energy of the particle is maximum at natural frequency. Since, the restoring force is proportional to displacement and resisting force is proportional to velocity. So the correct option is ${{\omega } _{0}}={{\omega } _{2}}\,\And \,{{\omega } _{1}}\ne\,{{\omega } _{0}}$

The rate of loss of mechanical energy is proportional to the damping force times velocity, or P = b * v^2. For a given initial displacement, the initial velocity is zero, but the damping force acts as the system moves. The damping coefficient b is the primary factor. Comparing the options, the smallest b value (5 g/s) leads to the smallest rate of energy loss.

$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{MB _H}}$
$\dfrac{60}{14}= 2\pi \sqrt{\dfrac{l}{M(MB _H)}}$
$\therefore \dfrac{7}{5} = \sqrt{B _H +B}{B _H} $ or $ B= \dfrac{24}{25}B _H$
Hence,
$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{M(B _H-B)}}= 2\pi \sqrt{\dfrac{l}{MB(1-24/25)}}$
$= 5\times 2\pi \sqrt{\dfrac{l}{2MB}} = 5 \times \dfrac{60}{10}= 30$
$\therefore f= \dfrac{60}{30} = 2$ vibrations/ min

$\omega = \sqrt {\dfrac{k}{m}-\dfrac{r^2}{4m^2}}= \sqrt{\dfrac{k}{m}}\sqrt{1-\dfrac{r^2}{4mk}}$
 $\approx \omega _o \left(1-\dfrac{r^2}{8mk}\right) \approx$ (1 - 1%)

Let original amplitude $=A$

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be

Amplitude is given by:

$A={{A} _{o}}{{e}^{-\alpha t}}$

Where, $A$ is amplitude at time t.

t is time

${{A} _{0}}$ is initial aplitude

$\alpha $ is constant

At t = 1s

$A=\dfrac{{{A} _{0}}}{2}$

So,

$ \dfrac{{{A} _{0}}}{2}={{A} _{0}}{{e}^{-\alpha }} $

$ {{e}^{-\alpha }}=\dfrac{1}{2} $

At t = 2s

$ A={{A} _{0}}{{e}^{-2\alpha }} $

$ A={{A} _{0}}{{(\dfrac{1}{2})}^{2}} $

$ A=\dfrac{{{A} _{0}}}{4} $

A(t) = A0 * exp(-kt). Given A(6) = A0/27, so exp(-6k) = 1/27 = (1/3)^3. Thus exp(-2k) = 1/3. At t=2, A(2) = A0 * exp(-2k) = A0/3.

Amplitude A(t) = A0 * exp(-kt). A(5) = 0.9 * A0, so exp(-5k) = 0.9. Energy E is proportional to A^2. E(10) = E0 * (A(10)/A0)^2 = E0 * (exp(-10k))^2 = E0 * (exp(-5k))^4 = E0 * (0.9)^4 = 0.6561 * E0. The closest option is 0.73, suggesting a potential calculation difference or rounding.

In forced oscillations, the amplitude A of a damped harmonic oscillator driven by a periodic force F is given by A = F / sqrt(m^2(omega^2 - omega_d^2)^2 + omega_d^2 b^2), where omega is the natural frequency, omega_d is the driving frequency, and b is the damping constant.

Amplitude follows A(n) = A0 * r^n. After 100 oscillations, A(100) = A0/3. After 200 oscillations, A(200) = A0 * (r^100)^2 = A0 * (1/3)^2 = A0/9.

For a damped forced oscillator, the amplitude resonance occurs at a driving frequency p_r = sqrt(omega_0^2 - b^2/2m^2), which is less than the natural frequency omega_0.

Energy E(t) = E0 * exp(-2kt). E(15) = 15, E0 = 45. 15 = 45 * exp(-2k * 15). 1/3 = exp(-30k). ln(1/3) = -30k. k = ln(3)/30.

$A = {A _0}{e^{ - kt}}$

For successive amplitudes $m = 1$
amplitude reduction factor

$=ln\left( \dfrac { { x } _{ 1 } }{ { x } _{ 2 } }  \right) =ln\left(

\dfrac { 10 }{ 1 }  \right) =ln10=2.3$
amplitude reduction factor $=\dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } $
$\Rightarrow \dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } =2.3$
squaring both sides
$\dfrac { 39.478{ \delta  }^{ 2 } }{ 1-{ \delta  }^{ 2 } } =5.29\\ \Rightarrow { \delta  }^{ 2 }=0.118\\ \Rightarrow \delta =0.344$

This is a forced oscillation problem. The steady-state solution for x(t) with a driving force F sin(t) involves terms of sin(t) and sin(omega_0 * t). Given omega_0 = 2, the solution is a combination of the driving frequency and natural frequency.

As we know that the amplitude of forced oscillation is given as

$A=\dfrac{F _0}{\sqrt{m^2(\omega^2-\omega _d^2)^2+\omega _d^2b^2}}$

Here we know that when oscillator is in resonance then,

$\omega=\omega _d$

so we have

$A=\dfrac{F _0}{\omega _d b}$

$F _0=20\,N$

$m = 600\, g$

$\omega=\sqrt{\dfrac km}$

$\omega=\sqrt{\dfrac{100}{0.6}}$

$\omega=12.9\,rad/sec$

Now we have

$A=\dfrac{20}{12.9\times 0.2}$

$A=7.7\,m$

For an adiabatic process, PV^gamma = constant. The restoring force for a small displacement x is F = -A * dP = -A * (gamma * P0 / V0) * (A * x). This leads to the SHM equation with omega^2 = (A^2 * gamma * P0) / (M * V0). Frequency f = omega / (2 * pi).

A(t) = A0 * exp(-kt). A(1) = A0/2, so exp(-k) = 1/2. A(3) = A0 * (exp(-k))^3 = A0 * (1/2)^3 = A0/8. Thus X = 8 = 2^3.

The equation of motion is mx'' + bx' + k*x = 0. The angular frequency of the damped oscillation is omega = sqrt(k/m - (b/2m)^2) = sqrt(k/m - b^2/(4m^2)).

A(5) = 0.9 * A0. A(15) = A0 * (exp(-5k))^3 = A0 * (0.9)^3 = 0.729 * A0. Thus alpha = 0.729.

Damping is caused by opposing force, which decreases the frequency.

Ans: C

The amplitude, a, at time $t$ is given by $a=a _0\;exp(-\,\alpha t)$

$a _{50}=a _0\;exp(-\alpha\times 50T)=0.80\;a _0$

where $T$ is the period of oscillation

$a _{150}=a _0\;exp(-a\times 150T)$

$=a _0\;(0.8)^3=0.512\,a _0$

Its is a damped oscillation, where amplitude of oscillation at time $t$ is given by $A = a _0e^{-\gamma t}$
where $a _0 = $ initial amplitude of oscillation
$\quad \gamma = $ damping constant
As per question, $\displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma100/v}\quad                    ...(i)$
(where $v$ is the frequency of oscillation)
and $A = a _0e^{-\gamma200/v} \quad                ...(ii)$
From $(i)$; $\quad \displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma\times100/v} \quad            ...(iii)$
Dividing equation $(ii)$ by $(iii)$, we have

$\quad \displaystyle\dfrac{A}{a _0(1/3)} = \displaystyle\dfrac{e^{-\gamma\times200/v}}{e^{-\gamma\times100/v}} = e^{-\gamma\times100/v} = \displaystyle\dfrac{1}{3}$

or $A = a _0\times\displaystyle\dfrac{1}{3}\times\displaystyle\dfrac{1}{3} = \displaystyle\dfrac{1}{9}a _0$

In reality, a spring won't oscillate forever. Frictional force will decrease the oscillation until eventually, the system is at rest.

Underdamped oscillation is practical because there always be resistive force present in reality which will try to make an oscillating body to lose its energy. This loss of energy makes the motion damped motion.

Dampers are found bridges to prevent them from swaying due to wind otherwise this motion can hamper the condition of a bridge.

The force on body oscillating in resistive medium is 

Using conservation of momentum and the properties of a spring-mass system, the center of mass velocity remains constant. The displacement expressions are derived from the relative motion of the two masses oscillating about the center of mass.

To and fro motion of a particle about mean position is called vibrational motion.

The time taken by vibrating body to complete one vibration is called time period. so our given statement is false.