Question

A uniform rod of length $50\text{cm}$ and mass $1\text{kg}$ is supported by two identical wires as shown in the figure. Where should a mass of $5\text{kg}$ be placed on the rod so that the same tuning fork may excite the wire on the left into $1^{st}$ fundamental vibrations and that on the right into its second overtone? $(\text{Take}g=10{\text{m/s}}^2)$

• A. $1\text{cm}$ from left end.
• B. $2\text{cm}$ from letf end.
• C. $1\text{cm}$ from right end.
• D. $2\text{cm}$ from right end.

Answer 1

The correct option is A $1\text{cm}$ from left end.


Given that,
Length of rod, $l=50\text{cm}$
Mass of rod, $m=1\text{kg}$
Mass of block, $m_b=5\text{kg}$
According to question,
$1$st fundamental of left wire = Second overtone of right wire
$f_{l1}=3f_{r1}$
$\Rightarrow \frac{v_1}{\text{/}{2l}_1}=3\frac{v_2}{\text{/}{2l}_2}$
$\Rightarrow \sqrt{\frac{T_1}{{\mu }_1}}=3\sqrt{\frac{T_2}{{\mu }_2}}$
[since $l_1=l_2$]
$\Rightarrow \sqrt{\frac{T_1}{T_2}}=3$ [as ${\mu }_1={\mu }_2$]
$\Rightarrow T_1=9T_2$

From FBD of rod,
$T_1+T_2=(5+1)\times g=6g$
$\Rightarrow 10T_2=6g$
$\Rightarrow T_2=\frac{6g}{10},T_1=\frac{54g}{10}$
Let the mass be attached at distance $x$ from end $A$.
Net torque about end $A$ must be zero.
$(\sum \overrightarrow{{\tau }_A})=0$
$5g\times x+1g\times \left(\frac{l}{2}\right)=T_{2}l$
$\Rightarrow 5gx+g\left(\frac{l}{2}\right)=\frac{6g}{10}l$
$\Rightarrow 5x=\frac{l}{10}$ or $x=\frac{l}{50}$
$\therefore x=\frac{50\text{cm}}{50}=1\text{cm}$