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?

a

The output power is

b

The Amplitude would decrease by

Explanation:

From the question we are told that

    The diameter of the steel wire is = 1.0mm

    The raduis of this steel wire is

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

Where is the mass per unit length of the wire

   This is mathematically evaluated as

Where a is the area of the the wire =

          is the density of steel with a generally value of

  So  

          is velocity of the wave

   This is mathematically evaluated as    

substituting  60Hz for f

  We have    

      is the amplitude with a given value of 0.50 cm

          v  is the linear velocity of the wave

  This is mathematically evaluated as    

Where T is the tension with a given value of

Substituting values into equation 1

Since the doubling of the frequency does not affect the amplitude and  from  equation one the output power  is   of the Amplitude, Then the Amplitude would decrease by

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary