Question

Two wires of same material of radius 'r' and '2r' respectively are welded together end to end. The combination is then used as a sonometer wire under tension 'T'. The joint is kept midway between the two bridges. The ratio of the number of loops formed in the wires such that the joint is a node is

• A. 1 : 5
• B. 1 : 2
• C. 1 : 4
• D. 1 : 3

Hint:

Linear mass density ($m$) scales with the square of the radius ($m \propto R^2$), meaning wave velocity drops linearly with a larger radius ($v \propto 1/R$). To keep a matching oscillation frequency, a lower velocity requires shorter loop lengths, which packs more loops into the same space ($p \propto R$). Since the radius doubles, the loop count must double too!

Answer 1

The Correct Option is B

Step 1: Set the matching conditions.
The two welded parts share the same tension and the same frequency, and each part has the same length. The joint is a node, so each part has a whole number of loops.

Step 2: Use the frequency formula.
For a wire of radius $r$, $n=\frac{p}{2lr}\sqrt{\frac{T}{\rho\pi}}$. The thinner wire has radius $r$, the thicker has $2r$.

Step 3: Equate frequencies.
\[ \frac{p_1}{r}=\frac{p_2}{2r} \]

Step 4: Find the ratio.
This gives $\frac{p_1}{p_2}=\frac{1}{2}$. \[ \boxed{1:2} \]