The amplitude of an oscillator decreases to 36.8% of its

Katherine Walls 2021-12-17 Answered
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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Expert Answer

eninsala06
Answered 2021-12-18 Author has 37 answers
The amplitude of a damped oscillator decreases exponentially with time:
A=A0et2τ
where τ is the time constant. Using AA0=0.368 and applying the logarithmic function yields.
τ=t2ln(0.368)=5s

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habbocowji
Answered 2021-12-19 Author has 22 answers

Answer:
τ=5s
Explanation:
When a vibrating body is damped. Its amplitude starts to decrease. This decrement is exponential. And it is given as follows:
X=X0et2τ
where, τ=Time Constant=?
X= Instantaneous value of amplitude
X0= Initial Value of amplitude
t=time interval=10s
The ratio of decrement is given as:
XX0=36.8%=0.368
therefore, using these values, we get:
XX0=0.368=e10s2τ
Taking natural log (ln) on both sides, we get:
ln(0.368)=10s2τ
τ=10s2ln(0.368)
τ=5s

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nick1337
Answered 2021-12-27 Author has 575 answers

Step 1
The expression for find value of time constant is
A=A0et/2τ
τ=t2ln(AA0)
Step 2
Substitute values in the above equation
τ=10.0s2ln(0.368)
=5.0s

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