Chapter 14, Problem 78A

To determine

To Rank: The frequencies of the oscillation in strings from largest to smallest.

Explanation of Solution

Given:

Length of strings:

  $\begin{array}{l}{L}_A=27\text{cm}\\ L_B=30\text{cm}\\ L_C=30\text{cm}\\ L_D=24\text{cm}\end{array}$

Number of anti-nodes:

  $\begin{array}{l}{n}_A=3\\ n_B=4\\ n_C=3\\ n_D=4\end{array}$

All the strings have same mass per unit length $\left(\mu \right)$ and tension $\left(T\right)$ .

Formula Used:

In third mode, the frequency of vibration of string is given by:

  $f_3=\frac{3}{2L}\sqrt{\frac{T}{\mu }}$

In fourth mode, the frequency of vibration of string is given by:

  $f_4=\frac{2}{L}\sqrt{\frac{T}{\mu }}$

Calculations:

String A and C are in mode 3. As $L_C>L_A,\therefore f_A>f_C$ .

String B and D are in mode 4. As $L_B>L_D,\therefore f_D>f_B$

String B and C have same length different modes. String has higher frequency is higher mode. Thus, $f_B>f_C$ .

From the above three comparison, the rank of frequencies of the oscillation in strings from largest to smallest is $f_D>f_B>f_A>f_C$ .

Conclusion:

  $f_D>f_B>f_A>f_C$