Question
A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50N. The frequency of the fourth harmonic of the fundamental is
Answer: Option C
:
C
Given mass =4g=4×10−3kg
length = 80 cm = 0.8 m
μ=masslength=4×10−30.8=0.5×10−2
given T = 50 N
Frequency of nth harmonic is given by fn=nv2L
v=√Tμ=√500.5×10−2
=√104=102ms−1
f4=4×1022×0.8
=0.25×103=250Hz
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:
C
Given mass =4g=4×10−3kg
length = 80 cm = 0.8 m
μ=masslength=4×10−30.8=0.5×10−2
given T = 50 N
Frequency of nth harmonic is given by fn=nv2L
v=√Tμ=√500.5×10−2
=√104=102ms−1
f4=4×1022×0.8
=0.25×103=250Hz
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