Question 239: Oscillations

Question

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ of original amplitude in 2 seconds. If its amplitude after 6 second become $\left(\frac{1}{n}\right)$ times the original amplitude, the value of $n$ is ( $n$ is non zero integer)

Accepted Answer

Step 1: Write amplitude decay relation
In general, for damped oscillation, amplitude decays exponentially:
$ A(t) = A_0 e^{-\gamma t} $
where $\gamma$ is the damping constant.
Step 2: Use given condition at $t = 2s$
It is given that amplitude becomes $	frac{1}{3}$ of original (thatÃ¢ÂÂs what Ã¢ÂÂ$\left(	frac{1}{3}ight)^{\mathrm{rd}}$Ã¢ÂÂ must mean: oneÃ¢ÂÂthird).
$ A(2) = \frac{A_0}{3} = A_0 e^{-2\gamma} $
Cancel $A_0$:
$ e^{-2\gamma} = \frac{1}{3} $
So,
$ \gamma = \frac{1}{2} \ln(3) $
Step 3: Condition at $t = 6s$
$ A(6) = A_0 e^{-6\gamma} $
Substitute $\gamma = \frac{1}{2} \ln 3$:
$ A(6) = A_0 e^{-6 \cdot 	frac{1}{2} \ln(3)} \ = A_0 e^{-3 \ln(3)} \ = A_0 \cdot e^{\ln(3^{-3})} \ = A_0 \cdot \frac{1}{27} $
Step 4: Express as $	frac{1}{n}$
So,
$ \frac{A(6)}{A_0} = \frac{1}{27} $
Thus $n = 27$.
Final Answer:
$ \boxed{27} $
Ã¢ÂÂ Correct Option:
D (27)

Answer

9

Answer

3

Answer

81

Answer

27

Oscillations

The amplitude of a damped oscillator becomes \left(\frac{1}{3}\right)^{\mathrm{rd}} of original amplitude in 2 seconds. If its amplitude after 6 second become \left(\frac{1}{n}\right) times the original amplitude, the value of n is ( n is non zero integer)

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