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 = 10 m s−2.
Figure

Answer 1

The Mass Should Be Placed at 5 cm.

Explanation:

Mass should be placed 5 cm far

Length of the rod = L = 40 cm = 0.4 m

Mass of rod m = 1.2 kg

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

Given that $f_1 = 2f_r$

So,  $\frac {1}{2l}\sqrt\frac {T_1}{m} = \frac {2}{2l}\sqrt\frac {T_r}{m}$

$\sqrt\frac {T_l}{T_r}$ = 2  

$\frac {T_l}{T_r}$ = 4 ...(1)

From the free body diagram-

$T_l + T_r$ = 60 N

$4T_r + T_r$ = 60 N

So, $T_r$ = 12 N and $T_l$ = 48 N

Now taking moment about point A,

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

x = 5 cm