91Ó°ÊÓ

Resonant frequency$f_1=420\mathrm{Hz}$

Resonant frequency$f_2=325\mathrm{Hz}$

Distance between two supports is $l=75\mathrm{cm}\mathrm{or}0.75\mathrm{m}$

Using the concept of resonant frequency and using the equation for the resonance wavelength, we can find the lowest resonant frequency. Using this frequency and wavelength, it is possible to find the velocity of the wave.

Formula:

The standing wave equation $kx=n\mathrm{Ï€}.........\left(1\right)$

The wavenumber,$k=\frac{2\mathrm{π}}{\mathrm{λ}}.....\left(2\right)$

The velocity of a body, $v=f\mathrm{λ}.......\left(3\right)$

For a fixed length, using equation (2) in equation (1), the resonant wavelength is given as:

$$\begin{gathered} \frac{2\mathrm{π}}{\mathrm{λ}}L=n\mathrm{π} \\ \mathrm{λ}=\frac{2\mathrm{L}}{\mathrm{n}}....\left(4\right) \end{gathered}$$

Then using equation (4) in equation (3), the resonant frequency can be found as:

$f=\frac{nv}{2L}$

Therefore, the nth frequency is given as:

$f_n=\frac{nv}{2L}$

And, the (n+1)th term of resonant frequency is given as:

$f_{(n+1)}=\frac{(n+1)v}{2L}$

Therefore,

The frequency difference between two successive pair of the harmonic frequencies is given as:

$$\begin{gathered} ∆f=f_{(n+1)}-f_n \\ =\frac{(n+1)v}{2L}-\frac{nv}{2L} \\ =\frac{v}{2L} \\ =f_1....\left(5\right) \end{gathered}$$

Given that the frequencies represent the consecutive frequencies, from the equation (5), we can write the lowest resonant frequency as:

$$\begin{gathered} f_{(n+1)}-f_n=420-315 \\ =105\mathrm{Hz} \end{gathered}$$

Therefore, the lowest possible resonant frequency is 105 Hz

The longest possible wavelength should be equal to the length which is twice the distance between the support. Therefore,

${\mathrm{λ}}_1=2L$

So, if $f_1$is the lowest possible frequency, using equation (3) we can write the speed of the waves is given as:

$$\begin{gathered} v=f_1×2L \\ =105×2××0.75 \\ =157.5\mathrm{m}/\mathrm{s} \\ ≈158\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of wave speed is 158 m/s