MCQ
The amplitude of a damped oscillator becomes half in one minute. The amplitude after $3$ minute will be $\frac{1}{X}$ times the original, where $X$ is
- A$2 \times 3$
- ✓${2^3}$
- C${3^2}$
- D$3 \times {2^2}$
✓
Answer
Correct option: B.
${2^3}$
b
(b) Amplitude of damped oscillator $A = {A_0}{e^{ - \lambda t}};\;\lambda \; = $constant, $t = time$
(b) Amplitude of damped oscillator $A = {A_0}{e^{ - \lambda t}};\;\lambda \; = $constant, $t = time$
For $t =1 min$. $\frac{{{A_0}}}{2} = {A_0}{e^{ - \lambda t}} \Rightarrow {e^\lambda } = 2$
For $t = 3 min$ $A = {A_0}{e^{ - \lambda \times 3}} = \frac{{{A_0}}}{{{{({e^\lambda })}^3}}} = \frac{{{A_0}}}{{{2^3}}}$ $ \Rightarrow X = {2^3}$
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