Question

If the amplitudes of a damped harmonic oscillator at times t=0, \(t_1\) and \(t_2\) are \(A_0\), \(A_1\) and \(A_2\) respectively, then the amplitude of the oscillator at a time of \((t_1+t_2)\) is

• A. \( \frac{A_0+A_1+A_2}{3} \)
• B. \( \frac{A_2A_0}{A_1} \)
• C. \( \frac{A_1A_0}{A_2} \)
• D. \( \frac{A_1A_2}{A_0} \)

Hint:

For any quantity that decays exponentially, such as amplitude in damped oscillations or population in radioactive decay, the value at a time \(t_1+t_2\) is related to the values at \(t_1\) and \(t_2\) by \(A(t_1+t_2) = A(t_1)A(t_2)/A_0\), where \(A_0\) is the initial value.

Answer 1

The Correct Option is D

The amplitude of a damped harmonic oscillator decreases exponentially with time.
The formula for the amplitude A at time t is \( A(t) = A_0 e^{-\gamma t} \), where \(A_0\) is the initial amplitude at t=0 and \(\gamma\) is the damping constant.
We are given the amplitudes at three different times:
At t=0: \( A(0) = A_0 e^0 = A_0 \). (This is consistent).
At t=\(t_1\): \( A_1 = A_0 e^{-\gamma t_1} \). (Equation 1)
At t=\(t_2\): \( A_2 = A_0 e^{-\gamma t_2} \). (Equation 2)
We need to find the amplitude at time \( t = t_1 + t_2 \), let's call it \(A_{1+2}\).
\( A_{1+2} = A_0 e^{-\gamma (t_1 + t_2)} \).
Using the property of exponents, we can write this as:
\( A_{1+2} = A_0 e^{-\gamma t_1} e^{-\gamma t_2} \).
From Equation 1, we have \( e^{-\gamma t_1} = \frac{A_1}{A_0} \).
From Equation 2, we have \( e^{-\gamma t_2} = \frac{A_2}{A_0} \).
Now, substitute these expressions back into the equation for \(A_{1+2}\).
\( A_{1+2} = A_0 \left( \frac{A_1}{A_0} \right) \left( \frac{A_2}{A_0} \right) \).
\( A_{1+2} = \frac{A_0 A_1 A_2}{A_0^2} = \frac{A_1 A_2}{A_0} \).