Question

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscillator is 57.0Hz and it is observed that the amplitude of the wave on the steel wire is 0.54cm What is the power output of the oscillator, assuming that the wave is not reflected back?

Answer 1

The power output of the oscillator is 0.350 watt.

Step-by-step explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

`P=(1)/(2)*\mu*\omega^2* a^2* v`

Put the value into the formula

`P=(1)/(2)* A*\rho*\omega^2* a^2*(√(T))/(\mu)`

`P=(1)/(2)*3.14*(0.0005)^2*7850*(2*\pi*57.0)^2*(0.54*10^(-2))^2*\sqrt{(5.7)/(3.14*(0.0005)^2*7850)}`

`P=0.350\ Watt`

Hence, The power output of the oscillator is 0.350 watt.