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प्रश्न
A piano wire weighing 6⋅00 g and having a length of 90⋅0 cm emits a fundamental frequency corresponding to the "Middle C" \[\left( \nu = 261 \cdot 63 Hz \right)\]. Find the tension in the wire.
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उत्तर
Given: Length of the piano wire (L) = 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (fo) = 261.63 Hz
Linear mass density, m = `6/90` gm/cm
\[= \left\{ \frac{6 \times {10}^{- 3}}{90 \times {10}^{- 2}} \right\} kg/m\]
\[ = \left( \frac{6}{900} \right) kg/m\]
= 6.67 × 10−3
\[\text{ Fundamental frequency, } f_o = \frac{1}{2L}\sqrt{\left( \frac{T}{m} \right)}\]
\[\Rightarrow 261 . 63 = \frac{1}{\left( 2 \times 0 . 09 \right)} \sqrt{\left\{ \frac{T \times 900}{6} \right\}}\]
\[\Rightarrow 0.18 \times 261.63 = \sqrt{150 T}\]
\[\Rightarrow 150 T = \left( 261.63 \times 0.18 \right)^2 \]
\[\Rightarrow T = \frac{\left( 261.63 \times 0.18 \right)^2}{150}\]
\[ = 1478.52 N \approx 1480 N\]
Hence, the tension in the piano wire is 1480 N.
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