Question

A wire of length l having tension T and radius r vibrates with fundamental frequency f. Another wire of the same metal with length 2l having tension 2T and radius 2r will vibrate with fundamental frequency:
\[\begin{align}
  & A.f \\
 & B.2f \\
 & C.\dfrac{f}{2\sqrt{2}} \\
 & D.\dfrac{f}{2}\sqrt{2} \\
\end{align}\]

Answer 1

Hint: The fundamental frequency or the natural frequency is the lowest possible frequency of a periodic wave function. It is also known as the lowest frequency sum, obtained during the superposition on the waves. We can solve this question using the laws of stretched strings.
Formula used:
$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

Complete step-by-step solution:
We know that the natural frequency is given by $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$ where, $T$ is the tension, $l$ is the length and $\mu$ is the mass per unit length.
Let $f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$, when, $T$ is the tension, $l$ is the length, $r$ is the radius of the wire and $\mu$ is the mass per unit length.
Clearly, $\mu=\dfrac{m}{l}=\dfrac{\rho\times \pi(r)^{2} l}{l}=\rho\times \pi(r)^{2}$
Similarly if , \[2T\] is the tension, \[2l\] is the length, \[2r\] is the radius of the wire
Then $\mu\prime$ which is the mass per unit length given as $\mu\prime=\dfrac{m}{l}=\dfrac{\rho\times \pi(2r)^{2}2 l}{2l}=4\rho\times\pi(r)^{2}$
thus, clearly, $4\mu= \mu\prime$
Then the new frequency will become $f’=\dfrac{1}{2(2l)}\sqrt{\dfrac{2T}{4\mu}}$
or, we get $f’ =\dfrac{f}{2\sqrt 2}$
Hence the answer is $C.\dfrac{f}{2\sqrt 2}$

Note: Mass per unit length is directly proportional to the square of the radius and is independent of the length of the wire. According to the laws of stretched strings gives the relationship between $f$ and the other components of the wave as follows:
1. the frequency of the wave is inversely proportional to the resonant length $L$.
2. the frequency of the wave is proportional to the square root of the tension $T$ experienced by the string.
3. the frequency id the wave is inversely proportional to the square root of the mass per unit length $\mu$ of the string.