A vibrating standing wave on a string radiates a sound wave with intensity proportional to the square of the standing-wave amplitude. When a piano key is struck and held down, so that the string continues to vibrate, the sound level decreases by 8.0 dB in 1.0 s. What is the string’s damping time constant t?

The concept of amplitude is maximum displacement or distance moved by a point on a waver about the point of equibrilium.

Provided information,

$\begin{array}{rcl}\beta & =& 8.0db\\ t& =& 1.0s\end{array}$

The solution to the equation is,

The vibration of the piano does not tend to be constant on the oscillating system that is responsible for decreasing the aptitude that changes over time. the gradual approach of the aptitude is 0. The expression shows the decay.

$A\left(t\right)=A\left(e^{-t/2r}\right)$

The A is amplitude, t is time and the time constant is ${}_T$

Converting level 8dB to the intensity

$$\begin{gathered} dB=10\log \left(\frac{I\left(t\right)}{I}\right) \\ 8dB=10\log \left(\frac{I\left(t\right)}{I}\right) \\ \left(\frac{I\left(t\right)}{I}\right)={10}^{-0.8} \\ \left(\frac{I\left(t\right)}{I}\right)={10}^{-0.8} \\ \left(\frac{I\left(t\right)}{I}\right)=0.158 \end{gathered}$$

On the other hand, $I\infty A^2,thenA-\sqrt{I}$

Therefore,

localid="1649070888565" role="math" $\begin{array}{rcl}\mathrm{A}\left(\mathrm{t}\right)& =& \left({\mathrm{e}}^{-\mathrm{t}/2\mathrm{{\rm T}}}\right)\\ \sqrt{\mathrm{I}\left(\mathrm{t}\right)}& =& \sqrt{\mathrm{I}\left({\mathrm{e}}^{-\mathrm{t}/2\mathrm{{\rm T}}}\right)}\\ \sqrt{0.0158}& =& \left({\mathrm{e}}^{-\mathrm{t}/2\mathrm{{\rm T}}}\right)\\ 0.398& =& {\mathrm{e}}^{-1.0/2\mathrm{{\rm T}}}\\ \mathrm{{\rm T}}& =& 0.543\mathrm{s}\\ & & \end{array}$