Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously with tensions ' T 1 ' and ' T 2 ' respectively. On changing the ten

Question

Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously with tensions ' T 1 ​ ' and ' T 2 ​ ' respectively. On changing the tension slightly in one of them, the beat frequency remains unchanged. This will happen when (Given → T 1 ​ > T 2 ​ )

TestHub · Accepted Answer

The frequency of vibration of a stretched string is given by $n = \frac{1}{2 L} \frac{T}{m}$ .

Thus, the frequency is directly proportional to the square root of tension, $n \propto T$ .

Given $T_{1} > T_{2}$ , we have $n_{1} > n_{2}$ .

The initial beat frequency is $n_{1} - n_{2} = 6$ .

If the tension in one of the wires is changed slightly and the beat frequency remains unchanged (i.e., 6 beats per second), the frequency of the changed wire must cross the frequency of the other wire to produce the same beat frequency on the opposite side.

Case 1: If tension $T_{1}$ is changed, the new frequency $n_{1}^{'}$ must satisfy $∣ n_{1}^{'} - n_{2} ∣ = 6$ .
Since $T_{1}$ is changed, $n_{1}^{'} \neq = n_{1}$ . Thus, $n_{2} - n_{1}^{'} = 6 \Rightarrow n_{1}^{'} = n_{2} - 6$ .
Since initially $n_{1} = n_{2} + 6$ , we get $n_{1}^{'} < n_{1}$ .
For the frequency to decrease, the tension $T_{1}$ must be decreased.

Case 2: If tension $T_{2}$ is changed, the new frequency $n_{2}^{'}$ must satisfy $∣ n_{1} - n_{2}^{'} ∣ = 6$ .
Since $T_{2}$ is changed, $n_{2}^{'} \neq = n_{2}$ . Thus, $n_{2}^{'} - n_{1} = 6 \Rightarrow n_{2}^{'} = n_{1} + 6$ .
Since initially $n_{2} = n_{1} - 6$ , we get $n_{2}^{'} > n_{2}$ .
For the frequency to increase, the tension $T_{2}$ must be increased.

Therefore, either $T_{1}$ is decreased or $T_{2}$ is increased.

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

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