c{1}{2L} \sqrt{\frac{T}{\mu}} $$

where $\mu$ is the mass per unit length of the wire.

Step 2: Frequency of closed pipe (fundamental and first overtone)

Length of closed pipe $= L$

Fundamental frequency of closed pipe:

$$ f_1 = \frac{v}{4L} $$

where $v$ is the speed of sound in air.
First overtone of closed pipe:

$$ f_2 = \frac{3v}{4L} $$

Step 3: Given resonance conditions

Case 1: Frequency of wire (with tension $T$) = fundamental freq. of pipe:

$$ \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L} $$
Case 2: Frequency of wire (with tension $T+8$) = first overtone of pipe:

$$ \frac{1}{2L} \sqrt{\frac{T+8}{\mu}} = \frac{3v}{4L} $$

Step 4: Relate the equations

Divide both sides of the first equation by $v/4L$:

$$ \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{v}{4L} $$

$$ \Rightarrow \sqrt{\frac{T}{\mu}} = \frac{v}{2} $$

Similarly, from the second equation:

$$ \frac{1}{2L} \sqrt{\frac{T+8}{\mu}} = \frac{3v}{4L} $$

$$ \Rightarrow \sqrt{\frac{T+8}{\mu}} = \frac{3v}{2} $$

Step 5: Form ratios to eliminate $v$ and $\mu$

Divide the second equation by the first:

$$ \frac{\sqrt{\frac{T+8}{\mu}}}{\sqrt{\frac{T}{\mu}}} = \frac{\frac{3v}{2}}{\frac{v}{2}} $$

$$ \frac{\sqrt{T+8}}{\sqrt{T}} = 3 $$

Square both sides:

$$ \frac{T+8}{T} = 9 $$

$$ T+8 = 9T $$

$$ 8 = 8T $$

$$ T = 1 \text{ N} $$

Final answer:

Option D, 1 N

Waves

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