Question

A long taut string is connected to a harmonic oscillator of frequency $f$ at one end. The oscillator oscillates with an amplitude $a_0$ and delivers power $P_0$ to the string. Due to dissipation of energy, the amplitude of wave goes on decreasing with distance $x$ from the oscillator as $a=a_0{e}^{-kx}.$ In what length of the string does ${\left(\frac{3}{4}\right)}^{th}$ of the energy supplied by the oscillator get dissipated?

• A. $\frac{\ln 2}{k}$
• B. $\frac{2\ln 2}{k}$
• C. $\frac{\ln 2}{2k}$
• D. $\frac{1}{k}$

Answer 1

The correct option is A $\frac{\ln 2}{k}$

Average power of a harmonic wave of amplitude $a_0$ on a string is given by
$P_0=\frac{1}{2}\mu {\omega }^2{a}_0^{2}v=2{\pi }^2\mu f^2{a}_0^{2}v$

At distance $x,a=a_0{e}^{-kx}$

$\therefore P=2{\pi }^2\mu f^2(a_0{e}^{-kx}{)}^{2}v$

Since $\frac{3}{4}$th of the supplied energy is dissipated, transmitted power: $P=\frac{P_0}{4}$
$\Rightarrow e^{-2kx}=\frac{1}{4}\Rightarrow -2kx=\ln \left(1\right)-\ln \left(4\right)\Rightarrow x=\frac{\ln 2}{k}$

$\Rightarrow x=\frac{\ln 2}{k}$

Thus, option (a) is the correct answer.