Strings Aand Bhave identical lengths and linear densities, but string Bis under greater tension than string A. Figure 16-27 shows four situations, (a) through (d), in which standing wave patterns exist on the two strings. In which situations is there the possibility that strings Aand Bare oscillating at the same resonant frequency?

Both strings A and B have identical lengths and linear densities

Identify the situation using the equations of frequency of the resonant standing wave and the velocity of the wave on the string.

Formulae are as follow:

$$\begin{gathered} f=n\frac{v}{2L} \\ v=\sqrt{\frac{\tau }{\mu }} \end{gathered}$$

Here, v is wave speed, T is tension in the string,𝝁 is mass per unit length,f is frequency, L is length.

The equation for the frequency of a standing wave on the string is,

$f=n\frac{v}{2L}$

Here, as the length is identical for both strings

$f\propto nv$

Now, the speed of the wave on the string is given by,

$v=\sqrt{\frac{\tau }{\mu }}$

Again, as the linear density for both strings is the same,

$v\propto \sqrt{\tau }$

So, substituting this is in the equation of frequency,

$f\propto n\sqrt{\tau }$

Also, the tension in string B is greater than the tension in string A.

So, to remain product $n\times \sqrt{t}$as constant, the number of harmonics of string A should be larger than that of string B.

This condition is satisfied only is situation‘d’. So, only in case‘d’, there is a possibility that both strings have the same resonant frequencies.

Hence, the strings A and B can possibly resonate at the same frequency in situation‘d’.

Therefore, determine the possible situation using the equations of frequency and velocity.