Question

An aluminium wire of length $l_1=60.0cm$ of cross-sectional area $1.00\times 10^{-2}c{m}^2$ is connected to a steel wire of the same cross-sectional area. The compound wire, loaded with a block $m$ of mass $10.0kg$ (as in fig), so that the distance $l_2$ from the joint to the supporting pulley is $86.6cm$. Transverse waves are set up in the wire by using an external source of variable frequency. What is the lowest frequency of excitation for which standing waves are observed such that the joint in the wire is a node?
(Take $g=1000cm/s^2$ ] $[{\rho }_{Al}=2.6g/c{m}^3$;
${\rho }_{\text{steel }}=7.8g/c{m}^3$ )

Answer 1

Answer: 326.8

Distance between two nodes $=\frac{\lambda }{2}$.
Let there are $P_1$ loops in Aluminium wire and $P_2$ loops in steel part of wire.
Allowed frequencies of Aluminium part:
$f_{Al}=\frac{P_1}{2l_1}\sqrt{\frac{T}{{\mu }_{Al}}}$ ...(i)
Allowed frequencies of steel part:
$f_{\text{steel }}=\frac{P_2}{2l_2}\sqrt{\frac{T}{{\mu }_{\text{steel }}}}$ ...(ii)
Area of croos-section $=1\times 10^{-2}c{m}^2$
${\mu }_{Al}=(2.6)\times \left(1\times 10^{-2}\right)=0.026g/cm$
$[\because \mu =\rho \times A]$
${\mu }_{\text{steel }}=(7.8)\times \left(1\times 10^{-2}\right)=0.078g/cm$
Equating (i) and (ii)
$\frac{P_1}{2\times 60}\sqrt{\frac{T}{0.026}}=\frac{P_2}{2\times 86.6}\sqrt{\frac{T}{0.078}}$
$\Rightarrow \frac{P_1}{P_2}=\frac{2\times 60\times 2}{2\times \sqrt{3}\times 100}\sqrt{\frac{T}{0.078}\times \frac{0.026}{T}}$
$[$ Take $86.6=\frac{\sqrt{3}}{2}\times 100]$
$=\frac{2\times 60}{\sqrt{3}\times 100\times \sqrt{3}}=\frac{2}{5}$
The lowest frequency of excitation causes
$P_1=2,P_2=5$

Lowest frequency $=\frac{2}{2\times 60}\sqrt{\frac{1000\times 10^4}{0.026}}$
$[\because T=Mg]$
$=\frac{10^5}{60\sqrt{26}}=\frac{10^5}{60\times 5.1}$
$=326.8Hz$