Question

A uniform horizontal rod of length $40cm$ and mass $1.2kg$ is supported by two identical wires as shown in the figure. Where should a mass of $4.8kg$ be placed on the rod from the left end (in cm) 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

$Lengthofrod=L=40cm=0.4mMassoftherodm=1.2Kg$

Let the $4.8Kg$ mass be placed at a distance x from the left end.

given that $f_1=2f_2\frac{1}{2L}\sqrt{\frac{T_1}{m}}=\frac{1}{2L}\times 2\times \sqrt{\frac{T_2}{m}}\sqrt{\frac{T_2}{T_2}}=\frac{2}{1}\therefore \frac{T_1}{T_2}=4\therefore T_1=4T_2$

now from the free body diagram

$T_1+T_2=48+12=60N4T_2+T_2=60N{T}_2=12N$and$T_1=48N$

Now taking moment about point A

$T_2\times 0.4=48x+12\left(0.2\right)\therefore x=0.05=5cm$

so the mass should be placed at a distance $5cm$ from the left end