A wire having a linear density 0.1kg/m is kept under a tension of 490 N. It is observed that it resonates at a frequency of 400Hz. The next higher frequency is 450 Hz. Find the length of the wire Share with your friends Share 18 Sanjay Singh answered this Dear student Formula of velocity on the string isv=Tμ=4900.1=4900=70m/sLet length of the wire be l.Let the wire vibrate forming n loops when it resonates with 400Hz.so l=nλ12...1v=f×λ170=400×λ1λ1=70400 l=nλ12=n2×70400=n2×740...2Let the wire vibrate forming (n+1 )loops when it resonates with 450Hz. l=(n+1)λ22...1v=f×λ270=450×λ2λ1=70450 l=(n+1)λ22=n2×70450=(n+1)2×745...3from eqn 2 and 3n2×740=(n+1)2×745n8=n+199n=8n+8n=8from eqn 3l=(8+1)2×745=92×745=710=0.7m=70cmRegards 52 View Full Answer