Damped Oscillations

A particle is oscillating freely with a natural frequency ${\omega }_0$ and amplitude $a$. It is later subjected to a damping force proportional to its velocity and keeps oscillating with a frequency $\omega .$ Which of the following statement is true?

A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to 

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude is $5 \mathrm{s}$. In another $10 \mathrm{s}$ it will decrease to $\mathrm{\alpha }$ times its original magnitude, where $\alpha$ equals. 

If a simple pendulum has significant amplitude (up to a factor of $\frac{1}{e}$ of original) only in the period between $t=0 \mathrm{s}$ to $t=\tau \mathrm{s}$ then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity, with $b$ as the constant of proportionality, the average lifetime of the pendulum is (assuming damping is small) in seconds

Damped harmonic oscillator consists of a block $\left(m=2 \mathrm{kg}\right),$ a spring $\left(k=8{\pi }^2 \mathrm{N}/\mathrm{m}\right),$ and a damping force $\left(F=-bv\right).$ Initially, it oscillates with an amplitude of $25 \mathrm{cm}.$ Because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. What is the value of $b$ (in ${10}^{-2} \mathrm{kg}/\mathrm{s}$)? (Assume small damping and take : $\ln \left(\frac{3}{4}\right)=-0.28$)

Which of the following differential equations represents a damped harmonic oscillator?

When a damped harmonic oscillator completes $100$ oscillations, its amplitude is reduced to ${\left(\frac{1}{3}\right)}^{\mathrm{rd} }$ of its initial value. What will be its amplitude when it completes $200$ oscillations?
 

$\mathrm{A}$ pendulum with time period of $1 \mathrm{s}$ is losing energy due to damping. At certain time its energy is $45\mathrm{J}.$ If after completing $15$ oscillations, its energy has become $15 \mathrm{J},$ its damping constant (in ${\mathrm{s}}^{-1}$ ) is

A damped oscillator consists of a spring-mass system with mass $2 \mathrm{kg}$ and spring of spring constant $10 \mathrm{N} {\mathrm{m}}^{-1}$. The damping force is given by $F=-b\left(\frac{\mathrm{d}x}{dt}\right)$ where $b=280 \mathrm{g} {\mathrm{s}}^{-1}.$ The time required for the amplitude of the oscillations to reduce to one-fourth ${\left(\frac{1}{4}\right)}^{th}$ of its initial value is: (Assume $\ln 2=0.7$)

The amplitude of a damped oscillator becomes half in one minute. The amplitude after $3$ minute will be $1/x$ times the original, where $x$ is

A damped harmonic oscillator has amplitude $16 \mathrm{cm}$ at $t=2 \sec$. Then, the amplitude of same oscillator at $t=8 \sec$ will be... (initial amplitude$=32 \mathrm{cm}$),

If equation of displacement of a damped oscillation is given by $x=e^{-0.2t}\sin (\omega t+\varphi )$ then time after which amplitude will be one fourth of its initial value-

Amplitude of an oscillating body become half after $20$ oscillations then after $100$ oscillations its amplitude become how much multiple of initial amplitude:

The amplitude of a damped oscillator becomes one third in $2 \sec .$ If its amplitude after $6 \sec$ is $1/n$ times the original amplitude then the value of $n$ is

An oscillator of mass $10 \mathrm{g}$ is oscillating with natural frequency of 100 Hz. Under slight damped conditions, a periodic force, $F=100\cos 20\pi t$ is applied on it. The amplitude of oscillation is approximately,

The amplitude of a S.H.M. reduces to $\frac{1}{3}$ in first $20 \mathrm{s}$. Then in first $40 \mathrm{s}$ its amplitude becomes -

A block of mass $200\mathrm{ }\mathrm{g}$ executing $\mathrm{S}\mathrm{H}\mathrm{M}$ under the influence of a spring of spring constant $\mathrm{k}=90\mathrm{ }\mathrm{N}\mathrm{ }{\mathrm{m}}^{-1}$ and a damping constant $\mathrm{b}=40\mathrm{ }\mathrm{g}\mathrm{ }{\mathrm{s}}^{-1}$ . What is the time elapsed for its amplitude to drop to half of its initial value ?(Given. In $(1/2)=-0.693$ )

When an oscillator completes $100$ oscillation its amplitude is reduced to $\frac{1}{3}$ of initial value. What will be its amplitude, when it completes $200$ oscillation: - 

The energy of the damped oscillator at any instant $t$ is given by, $E=E_0{\mathrm{e}}^{-\frac{\mathrm{bt}}{\mathrm{m}}}$, where $E_0$ is its initial energy and $b=40 \mathrm{g} {\mathrm{s}}^{-1}$ is the damping constant. For a block of mass, $m=200 \mathrm{g}$, find the time elapsed for its mechanical energy to drop to half of its initial value.