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प्रश्न
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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उत्तर
Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m)
\[= \left( \frac{4}{80} \right) g/cm = 0 . 005 kg/m\]
\[\text{ Wave speed, } \nu = \sqrt{\left( \frac{T}{m} \right)}\]
\[ = \sqrt{\left( \frac{50}{0 . 005} \right)} = 100 m/s\]
\[\text{ Fundamental frequency ,} f_o = \frac{1}{2L}\sqrt{\left( \frac{T}{m} \right)}\]
\[ = \frac{1}{2 \times 0 . 8} \times \sqrt{\left( \frac{50}{0 . 005} \right)}\]
\[ = \frac{100}{2 \times 0 . 8} = 62 . 5 Hz\]
\[\text { First harmonic = 62 . 5 Hz }\]
If f_4 =frequency of the fourth harmonic:
\[ \Rightarrow f_4 = 4 f_0 = 62 . 5 \times 4\]
\[ \Rightarrow f_4 = 250 Hz\]
\[\text{ Wavelength of thefourth harmonic,} \lambda_4 = \frac{\nu}{f_4} = \frac{100}{250}\]
\[ \Rightarrow \lambda_4 = 0 . 4 m = 40 cm\]
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