# The amplitude of an oscillator decreases to 36.8% of its

The amplitude of an oscillator decreases to 36.8% of its initial value in 10.0 s. What is the value of the time constant?
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eninsala06
The amplitude of a damped oscillator decreases exponentially with time:
$A={A}_{0}{e}^{-\frac{t}{2\tau }}$
where $\tau$ is the time constant. Using $\frac{A}{{A}_{0}}=0.368$ and applying the logarithmic function yields.
$\tau =-\frac{t}{2\mathrm{ln}\left(0.368\right)}=5s$

habbocowji

$\tau =5s$
Explanation:
When a vibrating body is damped. Its amplitude starts to decrease. This decrement is exponential. And it is given as follows:
$X={X}_{0}{e}^{-\frac{t}{2\tau }}$
where,
$X=$ Instantaneous value of amplitude
${X}_{0}=$ Initial Value of amplitude

The ratio of decrement is given as:
$\frac{X}{{X}_{0}}=36.8\mathrm{%}=0.368$
therefore, using these values, we get:
$\frac{X}{{X}_{0}}=0.368={e}^{\frac{10s}{2\tau }}$
Taking natural log (ln) on both sides, we get:
$\mathrm{ln}\left(0.368\right)=\frac{10s}{2\tau }$
$\tau =\frac{10s}{2\mathrm{ln}\left(0.368\right)}$
$\tau =5s$

nick1337

Step 1
The expression for find value of time constant is
$A={A}_{0}{e}^{-t/2\tau }$
$\tau =-\frac{t}{2\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}$
Step 2
Substitute values in the above equation
$\tau =-\frac{10.0s}{2\mathrm{ln}\left(0.368\right)}$
$=5.0s$