Tag: physics

Any oscillation in which the amplitude of the oscillating quantity decreases with time is termed as damped oscilaation

For a damped oscillator, amplitude A(t) = A0 * e^(-bt). Given A(6) = A0/27, we have e^(-6b) = 1/27, so e^(-2b) = (1/27)^(1/3) = 1/3. Thus, A(2) = A0 * e^(-2b) = A0/3.

A vibrometer is an instrument used to measure vibration, specifically displacement, velocity, or acceleration.

In damped oscillations, the relation of amplitude of oscillations with time is given by $y={ y } _{ o }{ e }^{ -bt }=\frac { { y } _{ o } }{ { e }^{ -bt } } $, where
${ y } _{ o }=$ initial amplitude of oscillation
$t=$time
$b=$damping constant
since $b>0\quad & \quad t>0;$
${ e }^{ bt }>1\ { \Rightarrow e }^{ -bt }<1\ { \Rightarrow { y } _{ o }e }^{ -bt }<{ y } _{ o }\ \Rightarrow y<{ y } _{ o }$
which means that as the time progress, its amplitude will decrease.

Mechanical energy E(t) = E0 * e^(-2bt), where b = c/(2m). Given m=0.4 kg, c=0.08 kg/s, b = 0.08 / (2 * 0.4) = 0.1 s^-1. We want E(t) = E0/2, so e^(-2 * 0.1 * t) = 0.5. -0.2t = ln(0.5) = -0.693. t = 0.693 / 0.2 = 3.465 s.

Amplitude A(t) = A0 * e^(-bt). Given A(t) = A0/2, e^(-bt) = 1/2. For 3t, A(3t) = A0 * (e^(-bt))^3 = A0 * (1/2)^3 = A0/8. Thus, n = 8.

Generally, work done by external force goes to total energy of the system. But in forced oscillations, it is dissipated by damping forces.

$\dfrac{A _0}{2} = A _0 e^{-0.1t} \Rightarrow e^{-0.1t} = 2 \Rightarrow 0.1t = \ell n 2$
$t = \dfrac{\ell n 2}{0.1} = 10 \, \ell n2 \approx 6.93 \approx 7s$

$f = 5$
so $T = \dfrac{1}{5}$

$10T = \dfrac{10}{5} = 2$

$\dfrac{A _0}{1000} = A _0 \left(\dfrac{1}{2}\right)^{t/2}$

$(2)^{t/2} = 1000$

$\left(\dfrac{t}{2}\right) log 2 = 3$

$t = \dfrac{6}{log 2} \approx 20 s$

It is our common experience that when a body is made to vibrate in a medium , the amplitude of the vibrating body continuously decreases with time and ultimately the body stops vibrating. This is called the damped vibrations.