
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
161.7k+ views
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.
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.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Class 11 JEE Main Physics Mock Test 2025

Differentiate between audible and inaudible sounds class 11 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
