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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. |
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Answer» Correct Answer - frequency = 250 and `lambda = 40cm` 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` |
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