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 re

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 ​ )

TestHub · Accepted Answer

The frequency of a stretched wire is given by $f = \frac{1}{2 L} \frac{T}{μ}$ .

Since $T_{1} > T_{2}$ , the initial frequencies satisfy $f_{1} > f_{2}$ .

The initial beat frequency is $f_{1} - f_{2} = 8$ .

For the beat frequency to remain unchanged after changing the tensions, the new frequencies must satisfy $∣ f_{1}^{'} - f_{2}^{'} ∣ = 8$ .

If $T_{1}$ is changed while $T_{2}$ is kept constant, we need $∣ f_{1}^{'} - f_{2} ∣ = 8$ . Since $f_{1} - f_{2} = 8$ , the only other possibility is $f_{2} - f_{1}^{'} = 8 \Rightarrow f_{1}^{'} = f_{2} - 8 = f_{1} - 16$ . This requires $f_{1}$ to decrease, meaning $T_{1}$ must be decreased.

If $T_{2}$ is changed while $T_{1}$ is kept constant, we need $∣ f_{1} - f_{2}^{'} ∣ = 8$ . Since $f_{1} - f_{2} = 8$ , the only other possibility is $f_{2}^{'} - f_{1} = 8 \Rightarrow f_{2}^{'} = f_{1} + 8 = f_{2} + 16$ . This requires $f_{2}$ to increase, meaning $T_{2}$ must be increased.

Thus, the beat frequency can remain unchanged if $T_{1}$ is decreased or $T_{2}$ is increased.

Answer: $T_{1}$ is decreased or $T_{2}$ is increased

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