(A) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-\left(\frac{b}{2m}\right)t}$.Given: $m = 200 \, g = 0.2 \, k

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(A) The amplitude of a damped harmonic oscillator at time $t$ is given by $A(t) = A_0 e^{-\left(\frac{b}{2m}\right)t}$.Given: $m = 200 \, g = 0.2 \, kg$,$b = 40 \, g \, s^{-1} = 0.04 \, kg \, s^{-1}$.We want to find $t$ when $A(t) = \frac{A_0}{2}$.Substituting the values: $\frac{A_0}{2} = A_0 e^{-\left(\frac{0.04}{2 \times 0.2}\right)t}$.$\frac{1}{2} = e^{-\left(\frac{0.04}{0.4}\right)t} = e^{-0.1t}$.Taking the natural logarithm on both sides: $\ln(0.5) = -0.1t$.$-0.693 = -0.1t$.$t = \frac{0.693}{0.1} = 6.93 \, s \approx 7 \, s$.

Class 11 Physics Oscillations Different types of oscillations (Free, Damped, Forced Oscillation and Resonance)

Medium

$A$ 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 $\ln \frac{1}{2} = -0.693$).

A
$7$
B
$9$
C
$4$
D
$11$

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Oscillations

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Different types of oscillations (Free, Damped, Forced Oscillation and Resonance)

The amplitude of a damped oscillator varies with time as $A(t) = A_0 \exp(-bt / 2m)$, where $b = 70 \text{ g/s}$ and $m = 200 \text{ g}$. How long does it take for the mechanical energy to drop to one-fourth of its initial value (in $s$)? (Take $\ln 2 = 0.7$)

MediumTS EAMCET 2022

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

Easy

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The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial value is $t$. The time taken for its mechanical energy to drop to $50 \%$ of its initial mechanical energy is

DifficultAP EAMCET 2023

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The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

MediumAP EAMCET 2024

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For a damped harmonic oscillator governed by the equation $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$, find the time $t$ after which the mechanical energy becomes half of its initial maximum value.

DifficultAIIMS 2019

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$A$ 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 $\ln \frac{1}{2} = -0.693$).

Similar Questions

The amplitude of a damped oscillator varies with time as $A(t) = A_0 \exp(-bt / 2m)$, where $b = 70 \text{ g/s}$ and $m = 200 \text{ g}$. How long does it take for the mechanical energy to drop to one-fourth of its initial value (in $s$)? (Take $\ln 2 = 0.7$)

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

The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial value is $t$. The time taken for its mechanical energy to drop to $50 \%$ of its initial mechanical energy is

The displacement of a damped oscillator is $x(t) = \exp(-0.2 t) \cos(3.2 t + \Phi)$,where $t$ is time in seconds. The time required for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is (in $s$)

For a damped harmonic oscillator governed by the equation $m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$, find the time $t$ after which the mechanical energy becomes half of its initial maximum value.

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