Wave Motion 2 Question 22
22. An object of specific gravity $\rho$ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is $300 Hz$. The object is immersed in water, so that one half of its volume is submerged. The new fundamental frequency (in $Hz$ ) is
(a) $300\left(\frac{2 \rho-1}{2 \rho}\right)^{1 / 2}$
(b) $300\left(\frac{2 \rho}{2 \rho-1}\right)^{1 / 2}$
(c) $300\left(\frac{2 \rho}{2 \rho-1}\right)$
(d) $300\left(\frac{2 \rho-1}{2 \rho}\right)$
(1995, 2M)
Show Answer
Answer:
Correct Answer: 22. (a)
Solution:
- The diagramatic representation of the given problem is shown in figure. The expression of fundamental frequency is
$$ \begin{aligned} & \text { In air } \\ & T=m g=(V \rho) g \\ & \therefore \quad v=\frac{1}{2 l} \sqrt{\frac{A \rho g}{\mu}} \end{aligned} $$
When the object is half immersed in water
$$ \begin{aligned} T^{\prime} & =m g-\text { upthrust }=V \rho g-\left(\frac{V}{2}\right) \rho _w g \\ & =\left(\frac{V}{2}\right) g\left(2 \rho-\rho _w\right) \end{aligned} $$
The new fundamental frequency is
$$ \begin{array}{rlrl} & v^{\prime} & =\frac{1}{2 l} \times \sqrt{\frac{T^{\prime}}{\mu}}=\frac{1}{2 l} \sqrt{\frac{(V g / 2)\left(2 \rho-\rho _w\right)}{\mu}} \\ \therefore & \frac{v^{\prime}}{v} & =\left(\sqrt{\left.\frac{2 \rho-\rho _w}{2 \rho}\right)}\right. \\ \text { or } \quad v^{\prime} & =v\left(\frac{2 \rho-\rho _w}{2 \rho}\right)^{1 / 2} \\ & =300\left(\frac{2 \rho-1}{2 \rho}\right)^{1 / 2} Hz \end{array} $$
