Question

A small steel wire of diameter 1.0 mm is connected to an oscillator and is under a tension of 7.5 N. The frequency of the oscillator is 60.0 Hz and it is observed that the amplitude of the wave on the steel wire is 0.50 cma.What is the power output of the oscillator, assuming that the wave is not reflected back?b.If the power output stays constant but the frequency is doubled, what is the amplitude of the wave?

Answer 1

Answer:

a

The output power is $P= 0.764Watt$

b

The Amplitude would decrease by $\frac{1}{2}$

Explanation:

From the question we are told that

    The diameter of the steel wire is = 1.0mm$= \frac{1}{1000} = 1.0*10^{-3}m$

    The raduis of this steel wire is $r = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Now from the question we can deduce that the power output is equal to the power being transmitted by wave on the wire  this is mathematically represented as

                   $P = \frac{1}{2} \mu w^2 A^2 v ----(1)$

Where $\mu$ is the mass per unit length of the wire

   This is mathematically evaluated as

                      $\mu = a* \rho$

Where a is the area of the the wire = $\pi r^2 = (3.142 * 0.5*10^{-3})^2 =7.855*10^{-7}m^2$

         $\rho$ is the density of steel with a generally value of $7850 kg/m^3$

  So  

        $\mu = 7.855*10^{-7} *7850$

           $= 6.162*10^{-3}kg/m$

          $w$ is velocity of the wave

   This is mathematically evaluated as    

                   $w=2 \pi f$

substituting  60Hz for f

  We have    

                   $w = 2 *3.142 * 60$

                      $=377.04 \ rad/s$

      $A$ is the amplitude with a given value of 0.50 cm $= \frac{0.50}{100} = 0.50 *10^{-2}m$

          v  is the linear velocity of the wave

  This is mathematically evaluated as    

                  $v = \sqrt{\frac{T}{\mu} }$

Where T is the tension with a given value of $7.5N$

                $v = \sqrt{\frac{7.5}{6.162*10^{-3}} }$

                  $=34.89 m/s$

Substituting values into equation 1

       $P = 6.162*10^{-3}* 377.04^2 * (0.5*10^{-2})^2 * 34.89$

           $P= 0.764Watt$

Since the doubling of the frequency does not affect the amplitude and  from  equation one the output power  is  $\ \frac{1}{2}$ of the Amplitude, Then the Amplitude would decrease by $\frac{1}{2}$