A wire of density 8 g/cm³ is stretched between clamps 80 cm apart with an extension of 0.04 cm. Y = 8 × 1 0 11 dyne/cm². Find the lowest transverse frequency. (answer in Hz ).

Strain = 5 × 1 0 − 4 , stress = 4 × 1 0 8 , v = 4 × 1 0 8 /8 = 7071 cm/s, f = 7071/160 = 44.2 Hz.

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A wire of density 8 g/cm³ is stretched between clamps 80 cm apart with an extension of 0.04 cm. $Y = 8 \times 1 0^{11}$ dyne/cm². Find the lowest transverse frequency. (answer in $Hz$ ).

Answer:

44.20

Solution:

Strain = $5 \times 1 0^{- 4}$ , stress = $4 \times 1 0^{8}$ , $v = 4 \times 1 0^{8} /8 = 7071$ cm/s, $f = 7071/160 = 44.2$ Hz.

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Waves

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