91影视

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:

$$\begin{gathered} f=n\frac{v}{2L} \\ v=\sqrt{\frac{蟿}{渭}} \end{gathered}$$

Here, v is wave speed, T is tension in the string,饾潄 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

$f鈭�nv$

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

$v=\sqrt{\frac{蟿}{渭}}$

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

$v鈭�\sqrt{蟿}$

So, substituting this is in the equation of frequency,

$f鈭�n\sqrt{蟿}$

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

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

This condition is satisfied only is situation鈥榙鈥�. So, only in case鈥榙鈥�, 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鈥榙鈥�.

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