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# A pendulum with a time period of 1 sec is losing energy due to damping. At a certain time its energy is 45J. If after completing 15 oscillations, its energy has become 15 J its damping constant (in ${s^{ - 1}}$) is:A) $\dfrac{1}{2}$ B) $\dfrac{1}{{15}}\ln 3$ C) $\dfrac{1}{{30}}\ln 3$ D) 2

Last updated date: 19th Jun 2024
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Hint: In case of pendulum oscillating with decreasing amplitude/damping oscillation due to a damping force being applied on it loses energy gradually due loss of amplitude at an exponential rate.

Formula Used:
${A_t} = {A_0}{e^{ - bt/2m}}$
${A_t}$ : Amplitude of oscillation at time ${t}$
${A_0}$ : Amplitude of oscillation at time ${t=0}$
$\dfrac{b}{m}$ : damping coefficient

Step-1:
A simple pendulum oscillating with initial amplitude of motion as ${A_o}$ is decreased in the presence of dissipative forces after time ‘t’ it is given as,
${A_t} = {A_0}{e^{ - bt/2m}}$ ………… (1)
Where $b/m$ is the damping constant.
Step-2:
Now the Initial energy of oscillation can be given as
${E_0} = \dfrac{1}{2}k{A_0}^2 = 45J$  (assuming complete energy in form of potential energy)
Here,   [K = constant]
And after time t =15 secs that is after 15 oscillations (as period of oscillation is 1 second) its energy will be
${E_t} = \dfrac{1}{2}K{({A_0}{e^{ - bt/2m}})^2} = 15J$
Using equation (1) in above relation ${A_t} = {A_0}{e^{ - bt/2m}}$
${E_t} = \dfrac{1}{2}K{A_0}^2{e^{ - 2bt/2m}}$
Submitting values now we have
$15 = 45{e^{ - bt/m}}$
$\Rightarrow \dfrac{1}{3} = {e^{ - 15b/m}}$
Taking logarithm both sides $\ln \dfrac{1}{3} = - 15\dfrac{b}{m}$
$- \ln (3) = - 15\dfrac{b}{m}$
Therefore, $\dfrac{b}{m} = \dfrac{{\ln (3)}}{{15}}$ is the answer.
Hence, option (B) is correct.