The amplitude of a mass-spring system,which i | vedclass

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

Question

(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 ampl

Vedclass · Accepted Answer

$34.65$

Vedclass · 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 $\ln 2 = 0.693$:$t = \frac{2 	imes 500}{20} 	imes 0.693 = 50 	imes 0.693 = 34.65 \, s$.

Similar Questions

In forced oscillation of a particle,the amplitude is maximum for a frequency $\omega_{1}$ of the driving force,while the energy is maximum for a frequency $\omega_{2}$ of the driving force. Then:

In case of damped oscillation,the frequency of oscillation is ............

If the amplitude of a lightly damped oscillator decreases by $1.5 \%$,then the mechanical energy of the oscillator lost in each cycle is: (in $\%$)

$A$ block of mass $200 \, g$ is executing $SHM$ under the influence of a spring with spring constant $K = 90 \, N \, m^{-1}$ and a damping constant $b = 40 \, g \, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $\ln \frac{1}{2} = -0.693$).

In forced oscillations,a particle oscillates simple harmonically with a frequency equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module