A long taut string is connected to a harmonic oscillator of frequency f at one end. The oscillator oscillates with an amplitude a0 and delivers power P0 to the string. Due to dissipation of energy, the amplitude of wave goes on decreasing with distance x from the oscillator as a=a0e−kx. In what length of the string does (34)th of the energy supplied by the oscillator get dissipated?

A long taut string is connected to a harmonic oscillator of frequency

1.	A long taut string is connected to a harmonic oscillator of frequency f at one end. The oscillator oscillates with an amplitude a0 and delivers power P0 to the string. Due to dissipation of energy, the amplitude of wave goes on decreasing with distance x from the oscillator as a=a0e−kx. In what length of the string does (34)th of the energy supplied by the oscillator get dissipated?
Answer» 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 $a = a_{0} e^{- k x} .$ In what length of the string does ${(\frac{3}{4})}^{t h}$ of the energy supplied by the oscillator get dissipated?

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