The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to $$\alpha$$ times its original magnitude, where $$\alpha$$ equals :

For a damped harmonic oscillator the displacement (and therefore the amplitude) decays exponentially with time. We write this mathematically as

$$A(t)=A_0\,e^{-\beta t},$$

where $$A_0$$ is the initial amplitude at $$t=0$$ and $$\beta$$ is the damping constant (in s−1).

We are told that after $$5\text{ s}$$ the amplitude becomes $$0.9$$ times its original value. Translating this into the equation we have

$$0.9\,A_0 = A_0\,e^{-\beta\,(5)}.$$

Cancelling the common factor $$A_0$$ on both sides yields

$$0.9 = e^{-5\beta}.$$

To isolate $$\beta$$ we take the natural logarithm on both sides (recall that $$\ln e^x = x$$):

$$\ln(0.9) = -5\beta.$$

Thus

$$\beta = -\frac{1}{5}\,\ln(0.9) = \frac{1}{5}\,\ln\!\left(\frac{1}{0.9}\right).$$

Now we must find the amplitude after “another 10 s,” meaning a total elapsed time of

$$t = 5\text{ s} + 10\text{ s} = 15\text{ s}.$$

Using the same decay formula, the amplitude at $$t = 15\text{ s}$$ is

$$A(15) = A_0\,e^{-\beta\,(15)}.$$

Instead of substituting the logarithmic value of $$\beta$$ directly, it is simpler to recognize a pattern. We know

$$e^{-5\beta} = 0.9.$$

Therefore

$$e^{-15\beta} = \bigl(e^{-5\beta}\bigr)^3 = (0.9)^3.$$

Carrying out the cube:

$$(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.729.$$

By definition, $$\alpha$$ is the factor by which the amplitude has decreased after those additional 10 s, i.e.

$$\alpha = e^{-15\beta} = 0.729.$$

Hence, the correct answer is Option A.