Question

An aluminium wire of length 60 cm is joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40n. Cross-sectional area is 1 mm2 (steel) and 3 mm2 (aluminium). The minimum frequency of the tuning fork which can produce standing waves with the joint as a node is

Answer 1

AN ALUMINIUM WIRE OF LENGTH 60cm is joined to STEEL WIRE OF LENGTH 80cm & STRETCHED BETWEEN 2 FIXED POINTS. THE TENSION IS 40N. THE CROSS SECTIONAL AREA OF THE STEEL WIRE IS 3 mm SQUARE & THAT OF ALUMINIUM WIRE IS 3 MM SQUARE. WHAT SHOULD BE THE MINIMUM FREQUENCY OF THE TUNING FORK WHICH CAN PRODUCE STANDING WAVES IN THE SYSTEM WITH JOINT BETWEEN WIRES AS NODE. THE DENSITY OF ALUMINIUM WIRE IS 2.6gm/cm Cube & THAT OF STEEL IS 7.6gm/cm CUBE


 

7 years ago

Answers : (1)

hiii archit first of all find mass per unit length as density  and cross sectional area is given ...now for aluminum wire it will be 7.8*10to the power -3 and for steel u will get same...just multuply area given to density.now if mass per unit length is same for both then their wave speed will be same .now we have to find min. frequency ,for min.frequency their should be max. wavelength and for max. wavelength their should be min. loops should be produced.....now max. distance of loop can be 20 cm. i hope u will understand this from figure.,provided by me,now value of wavelenght will be 20multiply 2=40cm.now we know that wave speed= whole underoot of T/M,HERE T IS tension and m is mass per unit length now by putting the values in given formula u will get wave speed.thats it now u can calculate freq, as u know frequency=wave speed /wavelength .remember put the wavelength in m i,e 0.4m,u will get approx. 180 .hopefully this will help u,if u have any doubt u can ask me..........gud luck