Question

Two identical straight wires are stretched so as to produce 8 beats per second when vibrating simultaneously with tensions \(T_{1}\) and \(T_{2}\) respectively. On changing the tensions, the beat frequency remains unchanged. This will happen when \((T_{1}>T_{2})\)

• A. both \(T_{1}\) and \(T_{2}\) is decreased by value 256 N
• B. both \(T_{1}\) and \(T_{2}\) is increased by value 256 N
• C. \(T_{1}\) is decreased or \(T_{2}\) is increased
• D. \(T_{1}\) is increased or \(T_{2}\) is decreased

Hint:

For identical wires the frequency is proportional to the square root of tension. The beat frequency stays the same only if the frequency gap stays equal to 8 Hz.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For identical wires \(f\propto\sqrt{T}\), so \(T_{1}>T_{2}\) means \(f_{1}>f_{2}\) and \(f_{1}-f_{2}=8\ \text{Hz}\).

Step 2: Condition
The beat frequency stays 8 if the gap is again 8. That happens if the wires swap order, so that \(f_{2}'-f_{1}'=8\).

Step 3: Which change
This needs \(f_{1}\) to fall or \(f_{2}\) to rise, so \(T_{1}\) is decreased or \(T_{2}\) is increased. Raising \(T_{1}\) or lowering \(T_{2}\) only widens the gap.

Final Answer:
The beat frequency can stay the same if \(T_{1}\) is decreased or \(T_{2}\) is increased. \[ \boxed{\text{(C) }T_{1}\text{ is decreased or }T_{2}\text{ is increased}} \]