Question

Figure (15-E10), shows an aluminum wire of length $60cm$, joined to a steel wire of length $80cm$ and stretched between two fixed supports. The tension produced is $40N$. The cross-sectional area of the steel wire is $1.0m{m}^2$ and that of the aluminum wire is $3.0m{m}^2$. 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 aluminum is $2.6gc{m}^{-3}$ and that of steel is $7.8gc{m}^{-3}$.

• A. $180Hz$
  
• B. $1800Hz$
• C. $170Hz$
• D. $90Hz$

Answer 1

The correct option is A $180Hz$

${\rho }_s=7.8g/c{m}^3,{\rho }_A=2.6g/c{m}^3$

$m_s={\rho }_s{A}_s=7.8\times {10}^{-2}g/cm$ (m = mass per unit length)

$m_A={\rho }_A{A}_A=2.6\times {10}^{-2}\times 3g/cm=7.8\times {10}^{-3}kg/m$

A node is always placed in the joint. Since aluminium and steel rod has same mass per unit length, velocity of wave in both of them is same.

$\Rightarrow v=\sqrt{T/m}\Rightarrow 500/7m/x$

For minimum frequency there would be maximum wavelength for maximum wavelength minimum no of loops are to be produced.

$\therefore$ maximum distance of a loop $=20cm$

$\Rightarrow$ wavelength $=\lambda =2\times 20=40cm=0.4m$

$\therefore f=v/\lambda =180Hz$