Question

A string that is stretched between fixed supports separated by 75.0 cmhas resonant frequencies of 420and 315 Hz , with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Short answer

1. The lowest resonant frequency is 105 Hz
2. The wave speed is 158 m/s

Step-by-step solution

Given data

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

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

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

Understanding the concept of resonant frequency  

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 $kx=n\mathrm{\pi }.........\left(1\right)$

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

The velocity of a body, $v=f\lambda .......\left(3\right)$

Step 3(a): Calculation of lowest resonant frequency

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

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

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

Step 4(b): Calculation of wave speed

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

${\lambda }_1=2L$

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

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

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