Question

A bouncing ball is an example of an anharmonic oscillator. If you Quadruple the maximum height, what happens to the period? (Assume that the ball keeps returning to the same height.)

Short answer

The time period of the bouncing ball is twice its initial value.

Step-by-step solution

Understanding concept of anharmonic oscillator

Here it is given that the bouncing ball is an example of an anharmonic oscillator. It means that the bouncing ball does not follow the rules of harmonic motion.

The expression for the time period in the case of the bouncing ball is given by

$T=\sqrt{\frac{2h}{g}}$

Here T is the time period (the sum of the time of ascent and the sum of the time of descent), h is the height and g is the acceleration due to gravity.

Finding solution

From the above equation, it is clear that the time period is directly proportional to the square root of the height of the bouncing ball.

$T\propto \sqrt{h}$...... (i)

It is given that the maximum height is quadrupled.

$h^{\text{'}}=4h$

$T^{\text{'}}\propto \sqrt{4h}$...... (ii)

Divide equation (ii) by equation (i).

$$\begin{gathered} \frac{T^{\text{'}}}{T}=\sqrt{\frac{4h}{h}} \\ =2 \end{gathered}$$

$\therefore T^{\text{'}}=2T$

Therefore, on quadrupling the maximum height, the time period becomes twice the original.