Question

A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (below). Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on left into its fundamental vibrations and that on right into its first overtone? Take $g = 10~m {s^{−2}}$.

Answer 1

Hint: In the given question, Length of the rod is 40 cm and two identical wire supporting the mass 1.2 kg. Length= 40cm, Mass=1.2kg.Let us assume mass of 4.8kg is placed at y distance from the left.

Formula used:
${f_1} = 2{f_2}$

Complete answer:
According to the question${f_1} = 2{f_2}$ , where ${f_1}$is the frequency on the left side and ${f_2}$ is the frequency on the right side.

Let us assume, ${T_1}$ and ${T_2}$ be the tension on the left and right side respectively
\[ \Rightarrow \frac{1}{{2L}}\sqrt {\frac{{{T_1}}}{m}} = \frac{1}{{2L}} \times 2 \times \sqrt {\frac{{{T_2}}}{m}} \]
\[ \Rightarrow \sqrt {\frac{{{T_1}}}{{T{}_2}}} = 2\]
\[ \Rightarrow \frac{{{T_1}}}{{{T_2}}} = 4 \cdots (1)\]

Now from the given question,
$ \Rightarrow {T_1} + {T_2} = 48 + 12 = 60N$
$ \Rightarrow 4{T_2} + {T_2} = 60N$
$ \Rightarrow {T_2} = 12N$
$ \Rightarrow {T_1} = 48N$

Now take look over point A
${T_2} \times 0.4 = 48y + 12(0.2)$
$\therefore y = 0.05 = 5cm$

Therefore, the mass must be placed 5cm behind the left end.

Note:In this type of question, particular attention must be taken while answering the problems and the proper values must be entered into the specified formula in order to arrive at the correct result with precision.