Question

A uniform horizontal rod of length $0.40\text{m}$ and mass $1.2\text{kg}$ is supported by two identical wires as shown in figure. Where a mass of $4.8\text{kg}$ should be placed from the left end of 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 $g=10{\text{m/s}}^2$)

• A. $5\text{cm}$
• B. $10\text{cm}$
• C. $15\text{cm}$
• D. $25\text{cm}$

Answer 1

The correct option is A $5\text{cm}$

Frequency of standing wave fixed at both ends is
$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

$f_1=f_2$

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

$T_1=4T_2---\left(i\right)$


${\overrightarrow{F}}_{net}=0$
${\overrightarrow{\tau }}_{net}=0$

${\overrightarrow{F}}_{net}=0$
$T_1+T_2=60---\left(ii\right)$
from$\left(i\right)\text{and}\left(ii\right)T_1=48N\text{and}{T}_2=12N$

${\overrightarrow{\tau }}_{net}=0$
torque about point shown is
$x\times 48+\frac{L}{2}\times 12=L{T}_2$
$x\times 48+\frac{0.4}{2}\times 12=0.4\times 12$
$⟹x=0.05m$
$x=5cm$