Animated Solution for Physics - Oscillations: The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\text{s}$. In another $10\text{s}$, it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals

Decoding Damped Oscillations

Imagine pushing a child on a swing and then stepping back. The swing doesn't stop immediately; it oscillates back and forth, but with each pass, it doesn't go quite as high. The air resistance and friction at the hinges slowly drain the energy from the system. This beautiful, gradual fading of motion is what physicists call a damped oscillation.

In the mathematical realm, the amplitude of a damped harmonic oscillator doesn't just decrease randomly; it follows a strict, elegant rule known as exponential decay. The amplitude $A(t)$ at any given time $t$ is governed by the master equation:

Here, $A_0$ is the initial amplitude (how high the swing started), and $\lambda$ is the damping constant (which depends on the friction and the mass of the system).

Our problem gives us a crucial piece of intel: after $5$ seconds, the amplitude has dropped to $0.9$ times its original magnitude. Let's plug this into our master equation.

At $t=5$, we have $A(5)=0.9A_0$. Substituting this yields:

The $A_0$ on both sides cancels out gracefully, leaving us with a powerful mathematical key:

We don't need to solve for $\lambda$ directly. In physics, it's often wiser to keep expressions intact rather than calculating messy decimals. This term, $e^{-5\lambda }$, is our "decay factor" for any $5$-second interval.

Now, the question asks for the amplitude in another $10$ seconds. This is where many students stumble. "Another $10$ seconds" means $10$ seconds after the first $5$ seconds have already passed.

Therefore, the total time elapsed from the very beginning is $t=5+10=15$ seconds. We need to find the amplitude at $t=15$, which the problem defines as $\alpha A_0$.

Let's write the amplitude equation for $t=15$:

How do we evaluate $e^{-15\lambda }$ without knowing $\lambda$? We use the magic of exponent rules! We know that $x^{ab}=(x^a{)}^b$. We can rewrite our exponent to utilize the key we found earlier:

This is the "Aha!" moment. We already know that $e^{-5\lambda }=0.9$. Let's substitute that right in:

Now, it's just simple arithmetic. $9^3=729$, so $(0.9{)}^3=0.729$.

Comparing this to the given form $\alpha A_0$, we can confidently conclude that $\alpha =0.729$.

This problem beautifully illustrates a fundamental property of exponential decay: over equal time intervals, the quantity decreases by a constant multiplicative factor.

In the first $5$ seconds, the amplitude was multiplied by $0.9$. In the next $5$ seconds (total $10$s), it gets multiplied by $0.9$ again. In the next $5$ seconds (total $15$s), it gets multiplied by $0.9$ a third time. Hence, $0.9\times 0.9\times 0.9=0.729$. Understanding this intuitive property allows you to solve such problems in seconds without writing a single equation!