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प्रश्न
Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1⋅0 mm2 and that of the aluminium wire is 3⋅0 mm2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminium is 2⋅6 g cm−3 and that of steel is 7⋅8 g cm−3.
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उत्तर
Given:
Length of the aluminium wire (La)= 60 cm = 0.60 m
Length of the steel wire (Ls)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (Aa) = 1.0 mm2
Area of cross-section of the steel wire (As) = 3.0 mm2
Density of aluminium (ρa) = 2⋅6 g cm−3
Density of steel (ρs) = 7⋅8 g cm−3
\[\text{ Mass per unit length of the steel, m_s } = \rho_s \times A_s \]
\[ = 7 . 8 \times {10}^{- 2} gm/cm\]
\[ = 7 . 8 \times {10}^{- 3} kg/m\]
\[\text{ Mass per unit length of the aluminium, m_A }= \rho_A A_A \]
\[ = 2 . 6 \times {10}^{- 2} \times 3 gm/cm\]
\[ = 7 . 8 \times {10}^{- 2} gm/cm\]
\[ = 7 . 8 \times {10}^{- 3} kg/m\]
A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
\[\Rightarrow v = \left( \frac{T}{m} \right)\]
\[ = \sqrt{\left\{ \frac{40}{7 . 8 \times {10}^{- 3}} \right\}}^- \]
\[ = \sqrt{\left( \frac{4 \times {10}^4}{7 . 8} \right)}\]
\[ \Rightarrow v = 71 . 6 m/s\]
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
\[\Rightarrow \text{ Wavelength, } \lambda = 2 \times 20 = 40 cm\]
\[\text{ Or } \lambda = 0 . 4 m\]
\[ \therefore Frequency, f = \frac{v}{\lambda} = \frac{71 . 6}{0 . 4} = 180 Hz\]
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