Question

A rod of length 0.3 $m$ having variable linear mass density from A to B as $\lambda ={\lambda }_{0}x$ (x is distance from A in meter), where ${\lambda }_0=100kg/m^2$ is suspended by two light wires of same length. Ratio of their linear mass density is 2 : 9. Then which of the following is/are correct ?

• A. Ratio of wave speed in wire - 1 to wire - 2 is 3 : 2
• B. Ratio of wave speed in wire - 1 to wire - 2 is 3 : 1
• C. Second harmonic in wire -1 has same frequency as third harmonic in wire -2
• D. Third overtone in wire -1 has same frequency as fifth overtone in wire -2

Answer 1

Answer: (A), (C), (D)

The correct options are
A Ratio of wave speed in wire - 1 to wire - 2 is 3 : 2
C Second harmonic in wire -1 has same frequency as third harmonic in wire -2
D Third overtone in wire -1 has same frequency as fifth overtone in wire -2

Given,
$\frac{dm}{dx}={\lambda }_{0}x$
$⟹dm={\lambda }_{0}xdx$
Centre of mass
$=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}=\frac{{\int }_0^L{\lambda }_0{x}^{2}dx}{{\int }_0^L{\lambda }_{0}xdx}=\frac{{\lambda }_0\frac{L^3}{3}}{{\lambda }_0\frac{L^2}{3}}=\frac{2L}{3}$
Mass of rod is,
${\int }_0^L{\lambda }_{0}xdx=\frac{{\lambda }_0{L}^2}{2}$
Apply force equilibrium on rod we get,
$T_1+T_2=\frac{{\lambda }_0{L}^2}{2}g$
Tourque equilibrium about point A on rod,
$T_2\times L=\frac{{\lambda }_0{L}^2}{2}g\times \frac{2L}{3}$
$T_2=\frac{{\lambda }_0{L}^{2}g}{3}$ $T_1=\frac{{\lambda }_0{L}^{2}g}{6}$
$L=0.3m$ ${\lambda }_0=100kg/m^2$
$T_2=30N,$ $T_1=15N$
Relation between wave speed and tension in wires,
$V_1=\sqrt{\frac{T_1}{{\mu }_1}}$ and $V_2=\sqrt{\frac{T_2}{{\mu }_2}}$
$\frac{V_1}{V_2}=\sqrt{\frac{T_1{\mu }_2}{T_2{\mu }_1}}=\frac{3}{2}$
n- harmonic frequency in a string is,
$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
Second harmonic of wire - 1,
$f_{21}=\frac{2}{2l}\sqrt{\frac{T_1}{{\mu }_1}}=\frac{2}{2l}{V}_1$
Third harmonic of wire - 2,
$f_{32}=\frac{3}{2l}\sqrt{\frac{T_2}{{\mu }_2}}=\frac{3}{2l}{V}_2$
$\frac{f_{21}}{f_{32}}=\frac{2\times 3}{3\times 2}=1$
We know, harmonic is one greater than overtone, so third overtone is equal to fourth harmonic and fifth overtone is equal to sixth harmonic.
hence, option (d) automatically get correct.