Question

The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges (from left end of sonometer) below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1:3:5
A. \[\dfrac{{1500}}{{23}}cm,\dfrac{{500}}{{23}}cm \\ \]
B. \[\dfrac{{1500}}{{23}}cm,\dfrac{{300}}{{23}}cm \\ \]
C. \[\dfrac{{300}}{{23}}cm,\dfrac{{800}}{{23}}cm \\ \]
D. \[\dfrac{{1500}}{{23}}cm,\dfrac{{2000}}{{23}}cm \]

Answer 1

Hint: When we put two bridges then we get three line segments. When all these three line segments are vibrating in fundamental overtones, then the wavelength of the wave generated will be twice the length of respective segments.

Formula used:
\[{f_n} = \dfrac{{nv}}{{2L}}\]
Where \[{f_n}\] is the frequency of nth order overtone, L is the length of the string, v is the speed of transverse wave on the string and n is the order of overtone.

Complete step by step solution:

Figure: The string with two bridges.

When both the ends are fixed, then the frequency of nth overtone is given as,
\[{f_n} = \dfrac{{nv}}{{2L}} \\ \]
For the fundamental frequency, the order of overtones is 1.
So, the fundamental frequency of the standing wave will be,
\[{f_0} = \dfrac{v}{{2L}} \\ \]
Let the positions of bridges be so that the whole length of the string is divided into segments of length \[{L_1},{L_2}\] and \[{L_3}\]. If the length of total string is L,
\[L = {L_1} + {L_2} + {L_3}\]
As the material of the string in each segment is the same, so the speed of the wave will be the same, let it be v. So, the fundamental frequencies are,
\[{f_{01}} = \dfrac{v}{{2{L_1}}} \\ \]
\[\Rightarrow {f_{02}} = \dfrac{v}{{2{L_2}}} \\ \]
\[\Rightarrow {f_{03}} = \dfrac{v}{{2{L_3}}}\]

It is given that the fundamental frequencies in each segment (from left to right) is in the ratio \[1:3:5\]
\[\dfrac{v}{{2{L_1}}}:\dfrac{v}{{2{L_2}}}:\dfrac{v}{{2{L_3}}} = 1:3:5 \\ \]
\[\dfrac{1}{{{L_1}}}:\dfrac{1}{{{L_2}}}:\dfrac{1}{{{L_3}}} = 1:3:5 \\ \]
\[\Rightarrow {L_1}:{L_2}:{L_3} = \dfrac{1}{1}:\dfrac{1}{3}:\dfrac{1}{5} \\ \]
\[\Rightarrow {L_1}:{L_2}:{L_3} = 15:5:3\]
If the factor is x, then
\[15x + 5x + 3x = L\]
\[\Rightarrow 23x = L\]
\[\Rightarrow x = \dfrac{L}{{23}} \\ \]
Hence, the lengths of the segments are,
\[{L_1} = \dfrac{{15L}}{{23}},{L_2} = \dfrac{{5L}}{{23}},{L_3} = \dfrac{{3L}}{{23}} \\ \]
So, the distance of first bridge from the left end is,
\[{x_1} = {L_1} = \dfrac{{15L}}{{23}} \\ \]
\[\Rightarrow {x_1} = \dfrac{{15 \times 100}}{{23}}\,cm = \dfrac{{1500\,cm}}{{23}}\]
And, the distance of second bridge from the left end is,
\[{x_2} = {L_1} + {L_2}\]
\[\Rightarrow {x_2} = \dfrac{{15L}}{{23}} + \dfrac{{5L}}{{23}} = \dfrac{{20L}}{{23}} \\ \]
\[\therefore {x_2} = \dfrac{{20 \times 100}}{{23}}cm = \dfrac{{2000cm}}{{23}}\]
Hence, the position of the two bridges from the left end is \[\dfrac{{1500}}{{23}}cm,\dfrac{{2000}}{{23}}cm\].

Therefore, the correct option is D.

Note: We should be careful about considering the bridge as the flexible end or the rigid end. Because based on the end of the string, the kind of waves generated changes.