Question:
A sonometer wire resonates with \(4\) antinodes between the 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 the bridges as before) The value of M is
A sonometer wire resonates with \(4\) antinodes between the 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 the bridges as before) The value of M is
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Frequency is proportional to p times sqrt(T).
Updated On: Oct 1, 2026
- \(2\) kg
- \(3\) kg
- \(4\) kg
- \(5\) kg
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For a sonometer wire with \(p\) loops (antinodes) between the bridges, the frequency is \(f=\dfrac p{2l}\sqrt{\dfrac T\mu}\). The tuning fork frequency, length and mass per length are the same in both cases.
Step 2: Relation:
So \(p\sqrt T\) is constant: \(4\sqrt{T_1}=2\sqrt{T_2}\).
Step 3: Solve:
\(\sqrt{T_2}=2\sqrt{T_1}\), so \(T_2=4T_1\). Tension is proportional to the suspended mass, so \(M=4\times1=4\) kg.
Step 4: Choose:
Option (C).
Final Answer:
M = 4 kg. \[ \boxed{4\ \text{kg}} \]
For a sonometer wire with \(p\) loops (antinodes) between the bridges, the frequency is \(f=\dfrac p{2l}\sqrt{\dfrac T\mu}\). The tuning fork frequency, length and mass per length are the same in both cases.
Step 2: Relation:
So \(p\sqrt T\) is constant: \(4\sqrt{T_1}=2\sqrt{T_2}\).
Step 3: Solve:
\(\sqrt{T_2}=2\sqrt{T_1}\), so \(T_2=4T_1\). Tension is proportional to the suspended mass, so \(M=4\times1=4\) kg.
Step 4: Choose:
Option (C).
Final Answer:
M = 4 kg. \[ \boxed{4\ \text{kg}} \]
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