A block of mass 100 ,g is connected to an ela | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 $\alpha = \frac{b}{2m} = \frac{69.3 \,g \,s^{-1}}{2 	imes 100 \,g} = \frac{69.3}{200} \,s^{-1} = 0.3465 \,s^{-1}$.Using $\ln 2 = 0.693$, we have $0.693 = \alpha t = 0.3465 	imes t$.Solving for $t$: $t = \frac{0.693}{0.3465} = 2 \,s$.

Similar Questions

In case of a forced vibration,the resonance wave becomes very sharp when the:

$A$ particle is performing simple harmonic motion. If the oscillations are damped oscillations,then the angular frequency is given by:

The amplitude of a damped harmonic oscillator becomes half in $3 \ s$ and will become $1/x$ of the initial amplitude in the next $6 \ s$,where $x$ is:

$A$ simple harmonic oscillator of angular frequency $\omega = 2 \, rad \, s^{-1}$ is acted upon by an external force $F = \sin(t) \, N$. If the oscillator is at rest in its equilibrium position at $t = 0$,its position at later times is proportional to

In the case of a forced vibration,the resonance curve becomes very sharp when the

Vedclass Test Series

Exam Paper Generator

Online Exam Module