Question

The length of the strings 1, 2, 3 and 4 are kept fixed at $\;{\mathbf{L}},{\text{ }}\dfrac{{{\mathbf{3L}}}}{{\mathbf{2}}}{\text{ }},\;\dfrac{{{\mathbf{5L}}}}{4}$and $\;\dfrac{{7{\mathbf{L}}}}{4}$ respectively. Strings 1,2,3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonics, respectively such that all the strings have the same frequency. The correct match for the tension in the four strings in the units of ${T_0}$ will be :
A. $I\; \to \;P,{\text{ }}II\; \to \;Q,{\text{ }}III\; \to \;R,{\text{ }}IV\; \to \;T$
B. $I\; \to \;P,{\text{ }}II\; \to \;Q,{\text{ }}III\; \to \;T,{\text{ }}IV\; \to \;U$
C. $I\; \to \;P,{\text{ }}II\; \to \;R,{\text{ }}III\; \to \;T,{\text{ }}IV\; \to \;U$
D. $I\; \to \;T,{\text{ }}II\; \to \;Q,{\text{ }}III\; \to \;R,{\text{ }}IV\; \to \;U$

Answer 1

It should be remembered that the musical instrument is made using four different metal strings 1,2 ,3 and 4 each with mass per unit length $\mu ,2\mu ,3\mu ,4\mu$ respectively. The four different magnitudes are as P = 1, Q = $\dfrac{1}{2}$ , R =$\dfrac{1}{{\sqrt 2 }}$ , S= $\dfrac{1}{{\sqrt 3 }}$ , T= $\dfrac{3}{{16}}$ ,U = $\dfrac{1}{{16}}$ . Also remember the representation of the four strings.

Complete step by step answer:
It is given that the length of all the strings are fixed at the given respective value. Now we are also told the harmonics they are vibrated at and also that all the strings have the same frequency. Hence in this problem we can easily equate all the given conditions using the formula of frequency at a given harmonic and find the required values.

Case 1: Length = $\;{\mathbf{L}}$ , Tension = ${T_0}$ , frequency = ${f_0}$
Thus ${f_1} = \dfrac{1}{{2{L_0}}}\sqrt {\dfrac{{To}}{\mu }}$ , hence we can find the value of $To$ and use in the next cases.

Case 2: Length = $\dfrac{{{\mathbf{3L}}}}{{\mathbf{2}}}{\text{ }}$ , Tension = ${T_2}$ , frequency = ${f_0}$
Thus ${f_2} = \dfrac{3}{{2 \times \dfrac{{3{L_0}}}{2}}}\sqrt {\dfrac{{{T_2}}}{{2\mu }}}$, now it is said that frequency is same, thus ${T_2} = \dfrac{{{T_0}}}{2}$ .Hence it matches with condition Q .

Case 3: Length = $\dfrac{{5{\mathbf{L}}}}{4}{\text{ }}$, Tension = ${T_3}$ , frequency = ${f_0}$
Thus ${f_3} = \dfrac{5}{{2 \times \dfrac{{5{L_0}}}{4}}}\sqrt {\dfrac{{{T_3}}}{{3\mu }}}$ , now it is said that frequency is same, thus ${T_3} = \dfrac{{3{T_0}}}{{16}}$ .Hence it matches with condition T.

Case 4: Length = $\dfrac{{7{\mathbf{L}}}}{4}{\text{ }}$, Tension = ${T_4}$ , frequency = ${f_0}$
Thus ${f_4} = \dfrac{{14}}{{2 \times \dfrac{{7{L_0}}}{4}}}\sqrt {\dfrac{{{T_4}}}{{4\mu }}}$ , now it is said that frequency is same, thus ${T_4} = \dfrac{{{T_0}}}{{16}}$ .Hence it matches with condition U.

Hence the correct answer is option C.

Note: Reading the question and understanding is important for these type of questions. The formula mentioned above should be remembered and the difference between harmonic and overtone should be understood.