Question

A wire, fixed at both ends, is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz.
a. What could be the maximum value of the fundamental frequency?
b. If transverse waves can travel on this string at a speed of \[40\,m{s^{ - 1}}\], what is its length?

Answer 1

Hint:The frequency at which the transfer of energy or the oscillation is the highest is known as the resonant frequency. At the resonant frequency, the amplitude is maximum.

Formula used:
The formula of wavelength is given as,
\[\lambda = \dfrac{v}{{{f_0}}}\]
Where, v is the velocity of the wave and $f_0$ is the fundamental frequency.

Complete step by step solution:
a. Fundamental frequency is the lowest frequency that any waveform can have. It is the lowest resonating frequency. Given that the wire is fixed at both ends with the resonant frequency of 240 Hz and 320 Hz, this means that the fundamental frequency will be having a value in between this. Therefore, the fundamental frequency can have a maximum value of,
\[{f_0} = 320 - 240 = 80\,Hz\]

Hence, the maximum value of fundamental frequency is $80\,Hz$.

b. In transverse waves, the particle displaces in a direction that is perpendicular to the direction in which the wave is propagating.Given that the speed with which the wave is travelling is \[40\,m{s^{ - 1}}\]. Suppose the length of the wave is \[\lambda \]. The wavelength is given by the formula
\[\lambda = \dfrac{v}{{{f_0}}}\]……(i)

Given that the two ends are fixed, that means it is a standing wave and length will be the integral multiple of half wavelengths in the string. Mathematically,
\[L = \dfrac{{n\lambda }}{2}\]
Here suppose, n=1. The above equation can be written as,
\[L = \dfrac{\lambda }{2}\]
\[\Rightarrow \lambda = 2L\]

From equation (i) it can be written that,
\[v = \lambda \times {f_0}\]
\[\Rightarrow v = 2L \times {f_0}\]
\[\Rightarrow L = \dfrac{v}{{2{f_0}}}\]
Substituting all the given values we get
\[L = \dfrac{{40}}{{2 \times 80}}\]
\[\Rightarrow L = \dfrac{1}{4}\]
$\therefore L=0.25\,m$

The length of the transverse wave will be 0.25 m.

Note:It is important to remember that the transverse waves are different from the longitudinal waves. As in transverse waves, the waves travel perpendicularly, but in the longitudinal wave, the particles of the wave simply oscillate back and forth about their mean position and do not actually travel.