Question

A sonometer wire resonates with 4 antinodes between two bridges for a given tuning fork, when 1 kg mass is suspended from the wire. Using same fork, when mass M is suspended, the wire resonates producing 2 antinodes between the two bridges (distance between two bridges is as before). The value of M is

• A. 2.5 kg
• B. 3.5 kg
• C. 4 kg
• D. 1 kg

Hint:

The number of loops is inversely proportional to the square root of tension ($p \propto 1/\sqrt{T}$). Halving the number of loops ($4 \rightarrow 2$) requires the internal tension to increase by a factor of $2^2 = 4$. Since the initial mass was 1 kg, the new mass must be $1 \times 4 = 4\ \text{kg}$ immediately!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A sonometer wire is excited by a fixed frequency tuning fork. We are given the number of loops/antinodes ($p$) formed under two different suspended mass tensions, and we need to determine the value of the second mass $M$.

Step 2: Detailed Explanation:
The fundamental resonant frequency $n$ of a stretched sonometer wire operating with $p$ antinodes across a fixed length $L$ is given by: $$ n = \frac{p}{2L}\sqrt{\frac{T}{m}} $$ where $T$ is the tension ($T = mg$) and $m$ is the linear mass density. Since both the frequency $n$ and wire length $L$ are kept constant, the parameters follow the proportional relation: $$ p\sqrt{T} = \text{constant} \implies p^2 T = \text{constant} $$ Setting up this inverse square proportionality relationship between the two states: $$ p_1^2 T_1 = p_2^2 T_2 \implies T_2 = T_1 \cdot \left(\frac{p_1}{p_2}\right)^2 $$ Substituting the given parameters ($p_1 = 4$, $T_1 = 1\ \text{kg-wt}$, and $p_2 = 2$): $$ M = 1 \cdot \left(\frac{4}{2}\right)^2 = 1 \cdot (2)^2 = 4\ \text{kg} $$

Step 3: Final Answer:
The value of the suspended mass M is 4 kg, which corresponds to option (C).