Question

A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.

Part A If the loudness drops to 80 % of its original value in 4.0 s , what is the time constant of the damped oscillation?

Part B How long does it take for the sound to be 20 % as loud as it was at the start?

Part C What fraction of the original loudness remains after 1.0 min?

Answer 1

(a)t = 4 s

Let Ao be the amplitude of intial loudness and later A(t) = 0.8 Ao

We know that, A(t) = Ao e-t/$\tau$ ; where t is the time and $\tau$ is the time constant.

0.8Ao = Ao e-t/$\tau$

0.8 = e^-t/$\tau$

Taking natural log both the sides we get

ln (0.8) = -t/$\tau$

-t/$\tau$ = -0.223

$\tau$ = t/0.223 = 4s/0.223 = 17.94 s

Hence, the time constant of the damped oscillation is = $\tau$ = 17.94 s.

(b)Now we have $\tau$ = 17.94 sec ; A(t) = 0.2 Ao

Again using the same relation

A(t) = Ao e-t/$\tau$

0.2Ao = Ao e-t/$\tau$

0.2= e^-t/$\tau$

Taking natural log both the sides we get

ln (0.2) = -t/$\tau$

-t/$\tau$ = -1.61

t = $\tau$ x 1.61 = 17.94 x 1.61 = 28.88 sec

Hence, t = 28.88 sec

(c)In this case we have $\tau$ = 17.94 sec; t = 1 min = 60 sec

A(t) = Ao e^-t/$\tau$

A = Ao x e^-60/17.94 = Ao e^-3.34 = 0.0354 Ao

Hence, A = 0.0354 Ao or Its 3.54 % of the original loudness.