A uniform horizontal rod of length 40 cm and mass 1⋅2 kg is supported by two identical wires as shown in figure. 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}.

#### Solution

Given:

Length of the rod (*L*) = 40 cm = 0.40 m

Mass of the rod (*m*) = 1.2 kg

Let the mass of 4.8 kg be placed at *x* distance from the left.

As per the question, frequency on the left side = *f*_{0}

Frequency on the right side = 2*f*_{0}

Let tension be *T*_{1} and *T*_{2}_{ }on the left and the right side, respectively.

\[\therefore \frac{1}{2L}\sqrt{\frac{T_1}{m}} = \frac{2}{2L}\sqrt{\frac{T_2}{m}}\]

\[ \Rightarrow \sqrt{\frac{T_1}{T_2}} = 2\]

\[ \Rightarrow \frac{T_1}{T_2} = 4 . . . (1)\]

From the free body diagram:

\[T_1 + T_2 = 48 + 12 = 60 N\]

\[ \Rightarrow 4 T_2 + T_2 = 5 T_2 = 60 N \left[ \text{ using equation } \left( 1 \right) \right]\]

\[ \therefore T_2 = 12 N\]

\[\text{ and } T_1 = 48 N\]

Now, taking moment about point A:

\[T_2 \times \left( 0 . 4 \right) = 48x + 12 \times 0 . 2\]

\[ \Rightarrow 4 . 8 = 48x - 2 . 4\]

\[ \Rightarrow 4 . 8x = 2 . 4\]

\[ \Rightarrow x = \frac{2 . 4}{4 . 8} = \frac{1}{20} m = 5 \text{ cm }\]

Therefore, the mass should be placed at a distance of 5 cm from the left end.