Question

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.

Answer 1

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 = f₀
Frequency on the right side = 2f₀
Let tension be T₁ and T₂ 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.