Question

A stretched wire of length 60 cm vibrates a fundamental note of frequency 256 Hz. If the length of the wire is reduced to 15 cm while the tension is kept constant, the fundamental frequency of the wire will be:
A) 64
B) 256
C) 512
D) 1024

Answer 1

Hint: A wave that travels on a stretched wire is a transverse wave. The frequencies of the vibrations of a wire are also waves. The frequency of waves on a string is inversely proportional to the length of the wire if the tension in the wire is constant.

Formula used: In this solution, we will use the following formula:
\[f = \dfrac{v}{{2L}}\] where $f$ is the frequency of the waves travelling on a string with velocity $v$ and length $L$
$v = \sqrt {\dfrac{T}{\mu }} $ where $v$ is the velocity of a wave travelling on a string with tension $T$ and linear mass density $\mu $.

Complete step by step answer
We’ve been given that a stretched wire has a fundamental note of frequency 256 Hz when it is 50 cm long and we want to find out the fundamental frequency when the length is cut down to 15 cm.
Using the formula of the frequency of waves on a string, we can write
\[f = \dfrac{v}{{2L}}\]
Where the velocity of a wave is given as
$v = \sqrt {\dfrac{T}{\mu }} $
Since we’ve been told the tension in the string remains constant, so will the velocity of the waves travelling on the string will also remain constant. So the frequency will only be affected by the change in length and since $f \propto \dfrac{1}{L}$, we have:
$\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{{L_2}}}{{{L_1}}}$
Substituting the value of ${f_1} = 256\,Hz$, ${L_1} = 60\,cm$and ${L_2} = 15\,cm$, we get
$\dfrac{{256}}{{{f_2}}} = \dfrac{{15}}{{60}}$
Solving for ${f_2}$, we get

${f_2} = 1024\,Hz$ which corresponds to option (D).

Note
Since we are still using the same wire, we must not worry about the linear mass density since it will remain constant for the same wire. However, this formula is only applicable for wires with constant linear mass density.