Question

A steel wire of length 40 cm and diameter 0.25 mm vibrate in unison with a tube open at both ends and of effective length 60 cm, when each is sounding its fundamental notes. The air temperature is 27 0 C. Find the tension in the wire. Given – velocity of sound in air at 0 0 C = 332 m/s and density of steel = 7800 kg/m 3 .

Answer 1

Answer:

Explanation:

Given - wire of length L=40cm=0.40m,m=4g=0.004kg,v=340m/s ,

frequency of second harmonic in wire of length L is given by ,

         f  

2

​  

=  

L

1

​  

 

M

T

​  

 

​  

 ,

where T= tension in wire ,

          M=m/L=0.004/0.40=0.01kg/m  , (mass per unit length)  ,

so      f  

2

​  

=1/0.40  

T/0.01

​  

=25  

T

​  

 ,

now fundamental frequency of air column ,

         f  

1

′

​  

=v/4l ,

where l=1m  ( length of air column) ,

hence , f  

1

′

​  

=340/(4×1)=85Hz ,

as vibrating wire sets vibration in air column, therefore   f  

2

​  

=f  

1

′

​  

 ,

          25  

T

​  

=85 ,

or        T=85  

2

/25  

2

=11.56N