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(D) Given that, the fundamental frequency of the first wire is $f_1 = 500 \,Hz$. Let the frequency of the second wire be $f_2$. The beat frequency is

Vedclass · Accepted Answer

$1.02$

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(D) Given that, the fundamental frequency of the first wire is $f_1 = 500 \,Hz$. Let the frequency of the second wire be $f_2$. The beat frequency is $f_b = 5 \,Hz$. When the tension in the first wire is decreased, its frequency decreases. Since it produces $5$ beats per second with the second wire, the frequency of the first wire must have been higher than the second wire initially, or the second wire is higher. Given the tension in the first wire is decreased, the new frequency $f_1' = f_1 - 5 = 495 \,Hz$. However, the standard interpretation for this problem is that the second wire has a fixed frequency $f_2 = 505 \,Hz$ (or $495 \,Hz$). Using the relation $f \propto \sqrt{T}$, we have $\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$. Taking $f_2 = 505 \,Hz$ and $f_1 = 500 \,Hz$, we get $\frac{T_2}{T_1} = (\frac{505}{500})^2 = (1.01)^2 = 1.0201 \approx 1.02$.

$A$ wire vibrates at a fundamental frequency of $500 \,Hz$. $A$ second identical wire produces $5$ beats per second with it when the tension in the first wire is slightly decreased. The ratio of the tension in the second wire to the tension in the first wire is approximately equal to

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