Animated Solution for Physics - Oscillations: A damped harmonic oscillator has a frequency of $5$ oscillations per second. The amplitude drops to half its value for every $10$ oscillations. The time it will take to drop to $1/1000$ of the original amplitude is close to

The problem of a damped harmonic oscillator is a classic example of how real-world physics deviates from ideal, frictionless models. In an ideal simple harmonic oscillator, a pendulum or a spring would swing forever. But in reality, forces like air resistance or internal friction sap energy from the system, causing the amplitude of oscillation to decay over time.

For a lightly damped harmonic oscillator, the amplitude doesn't just drop randomly; it follows a strict mathematical rule—an exponential decay. The amplitude $A$ at any time $t$ is given by the master equation: Here, $A_0$ is the initial amplitude, and $\gamma$ is the damping constant, which dictates how quickly the oscillations die out. Our goal in this problem is to first find this elusive $\gamma$, and then use it to predict the future state of the oscillator.

Finding the Damping Constant

The problem gives us a crucial clue: the frequency of the oscillator is $5\text{ Hz}$ (5 oscillations per second). This means the time period for one single oscillation is:

We are told that the amplitude drops to half its initial value ($A_0/2$) after exactly 10 oscillations. How much time is that?

Now, we plug this information into our decay equation. At $t=2\text{ s}$, the amplitude is $A_0/2$:

The $A_0$ terms cancel out beautifully, leaving us with:

Taking the natural logarithm ($\ln$) on both sides, we can extract $\gamma$: We now have the exact rate of decay for our specific oscillator!

Predicting the Future

The final question asks: how long will it take for the amplitude to drop to a mere $1/1000$ of its original value? We set up our decay equation one more time, but now we know $\gamma$:

Again, $A_0$ cancels out. We invert both sides to get rid of the negative exponent:

Taking the natural logarithm once more:

Since $1000=10^3$, we can use the power rule of logarithms ($\ln (x^y)=y\ln x$) to simplify this to:

Now, we substitute the value of $\gamma$ we found earlier:

Solving for $t$, we get our final expression:

To get the numerical value, we plug in the standard values for the natural logarithms ($\ln 10\approx 2.303$ and $\ln 2\approx 0.693$):

This value is incredibly close to $20\text{ s}$, making option (a) the correct choice. The beauty of this problem lies in how the exponential function perfectly models the gradual, inevitable loss of energy in a physical system.