A steel string 2 m long is under a strain of 2% ( ρ = 7.8 × 1 0 3 kg/m³, Y = 2 × 1 0 11 N/m²). Find the fundamental frequency. (answer in Hz ).

Stress = 4 × 1 0 9 N/m², v = 4 × 1 0 9 /7800 = 716 m/s, f = 716/4 = 179 Hz.

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A steel string 2 m long is under a strain of 2% ( $ρ = 7.8 \times 1 0^{3}$ kg/m³, $Y = 2 \times 1 0^{11}$ N/m²). Find the fundamental frequency. (answer in $Hz$ ).

Answer:

179

Solution:

Stress = $4 \times 1 0^{9}$ N/m², $v = 4 \times 1 0^{9} /7800 = 716$ m/s, $f = 716/4 = 179$ Hz.

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Waves

⏱ 2mℹ️ Source: QB

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