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The amplitude of a S.H.O. reduces to 1/3...

The amplitude of a S.H.O. reduces to 1/3 in first 20 secs. then in first 40 sec. its amplitude becomes -

A

`1/3`

B

`1/9`

C

`1/27`

D

`1/sqrt3`

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The correct Answer is:
To solve the problem step by step, we will use the formula for the amplitude of damped oscillations in simple harmonic motion (SHM). The amplitude of a damped oscillator at time \( t \) is given by: \[ x(t) = x_0 e^{-\lambda t} \] where: - \( x(t) \) is the amplitude at time \( t \), - \( x_0 \) is the initial amplitude, - \( \lambda \) is the damping constant, - \( t \) is the time. ### Step 1: Determine the amplitude after 20 seconds According to the problem, the amplitude reduces to \( \frac{1}{3} \) of its original value after 20 seconds. This can be expressed as: \[ x(20) = \frac{x_0}{3} \] Using the formula for damped oscillations, we can set up the equation: \[ \frac{x_0}{3} = x_0 e^{-\lambda \cdot 20} \] ### Step 2: Simplify the equation We can cancel \( x_0 \) from both sides (assuming \( x_0 \neq 0 \)): \[ \frac{1}{3} = e^{-\lambda \cdot 20} \] ### Step 3: Take the natural logarithm Taking the natural logarithm of both sides gives: \[ \ln\left(\frac{1}{3}\right) = -\lambda \cdot 20 \] ### Step 4: Solve for \( \lambda \) Rearranging the equation to solve for \( \lambda \): \[ \lambda = -\frac{\ln\left(\frac{1}{3}\right)}{20} \] ### Step 5: Determine the amplitude after 40 seconds Now we need to find the amplitude after 40 seconds. We use the same formula: \[ x(40) = x_0 e^{-\lambda \cdot 40} \] Substituting \( \lambda \): \[ x(40) = x_0 e^{-40 \cdot \left(-\frac{\ln\left(\frac{1}{3}\right)}{20}\right)} \] ### Step 6: Simplify the expression This simplifies to: \[ x(40) = x_0 e^{2 \ln\left(\frac{1}{3}\right)} \] Using the property of logarithms \( e^{\ln(a^b)} = a^b \): \[ x(40) = x_0 \left(\frac{1}{3}\right)^2 \] ### Step 7: Final result Thus, the amplitude after 40 seconds is: \[ x(40) = \frac{x_0}{9} \] ### Conclusion The amplitude of the simple harmonic oscillator after 40 seconds is \( \frac{x_0}{9} \). ---

To solve the problem step by step, we will use the formula for the amplitude of damped oscillations in simple harmonic motion (SHM). The amplitude of a damped oscillator at time \( t \) is given by: \[ x(t) = x_0 e^{-\lambda t} \] where: - \( x(t) \) is the amplitude at time \( t \), ...
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