Question

A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15 - E9). 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 $g=10m{s}^{-2}$

Answer 1

Length of the rod,

L = 40 cm = 0.4 m

mass of the rod = m = 1.2 kg

Let the 4.8 kg mass be placed at a distance 'x' from the left end.

Given that, $f_1=2f_2$

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

$\Rightarrow \sqrt{\frac{T_1}{T_2}}=2$

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

From the free body diagram

$T_1+T_2=48+12=60N$

$\Rightarrow 4T_r+T_r=5T_r=60N$

$\therefore T_r=12N$

and $T_1=48N$

Now taking moment about point A

$T_r\times \left(0.4\right)=48x+12\left(0.2\right)$

$\Rightarrow 4.8=48x-2.4$

$\Rightarrow 4.8x=2.4$

$\Rightarrow x=\frac{24}{48}=\frac{1}{20}m=5cm$

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