A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.

Answer

frequency = 250 and `lambda = 40cm`

Explanation

m=`(4/80)g/(cm)`
`=0.005(kg)/m`
`T=50N`
`L=80cm=0.8m`
`v=sqrt((T/m))`
`=sqrt((50/0.005))=100m/s`
Fundamental frequency
`=f=1/(2L)sqrt((T/m))`
`={1/(2xx0.8)xxsqrt((50/0.005))}`
`=100/((2xx0.8))`
`=100/1.6=62.Hz`
First `=62.5Hz`
`f_4`=Frequency of forth harmonic
`=4f_0=f_3=62.5xx4`
`=250Hz`
`v=f_4 lambda_4`
`rarr (v/f)=100/250`
`=0.4m=40cm`