Question 275: Waves

Question

The frequency of a stretched uniform wire of length $L$ under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by 8 N , it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is

Accepted Answer

Let the initial tension in the wire be $T$.
Step 1: Frequency of the wire (fundamental mode)
The fundamental frequency of a stretched wire of length $L$ is:
$$ f_{	ext{wire}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$
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:
$$ \frac{T+8}{T} = 9 $$
$$ T+8 = 9T $$
$$ 8 = 8T $$
$$ T = 1 	ext{ N} $$
Final answer:
Option D, 1 N

Answer

4 N

Answer

$\quad \frac{1}{2} \mathrm{~N}$

Answer

2 N

Answer

1 N

Waves

The frequency of a stretched uniform wire of length L under tension is in resonance with the fundamental frequency of a closed pipe of same length. If the tension in the wire is increased by 8 N , it is in resonance with the first overtone of the same closed pipe. The initial tension in the wire is

Choose the correct answer: