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Chapter 14: Problem 27

The amplitude of an oscillator decreases to \(36.8 \%\) of its initial value in \(10.0 \mathrm{s}\). What is the value of the time constant?

Short Answer

Expert verified
\(\tau = -10.0 s / \ln(0.368)\) seconds. It is a positive value.

Step by step solution

01

Understanding the given information

It is given that the amplitude of an oscillator decreases to \(36.8\%\) of its initial value in \(10.0 s\). Here, the final amplitude \(A\) is \(36.8\%\) of the initial amplitude \(A_0\), i.e., \(A = 0.368A_0\). The time \(t\) taken for this decrease is given as \(10.0 s\). The task is to find the value of the time constant \(\tau\).
02

Using the amplitude decay formula

We use the equation \(A = A_0 e^{-t/\tau}\). From the given information, we know the relationship between \(A\) and \(A_0\) (i.e., \(A = 0.368A_0\)) and that \(t = 10.0 s\). Substitute the known values into the equation: \(0.368A_0 = A_0 e^{-10.0 s/\tau}\). Then, cancel \(A_0\) from both sides of the equation.
03

Solving for the time constant

After cancelling out the \(A_0\) term in the last step equation, we get \(0.368 = e^{-10.0 s/\tau}\). This equation needs to be solved for \(\tau\). First, take the natural logarithm of both sides which gives: \(\ln(0.368) = -10.0 s / \tau\). Then, multiply both sides by \(-\tau\), yielding \(-\tau \ln(0.368) = 10.0 s\). Finally, solve for \(\tau\) to get the time constant.

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