
Two strings A and B, made of the same material, are stretched by the same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed \[{v_A}\] and on B with speed \[{v_B}\] . The ratio \[\dfrac{{{v_A}}}{{{v_B}}}\] is
A. \[\dfrac{1}{2} \\ \]
B. 2
C. \[\dfrac{1}{4} \\ \]
D. 4
Answer
162.9k+ views
Hint:Using the relation given for the radius of strings, we find the relation for the mass per unit length of both the strings. As both are made of the same material, the density should be the same.
Formula used:
\[v = \sqrt {\dfrac{T}{\mu }} \]
where v is the speed of the transverse wave on a string with \[\mu \] as the mass per unit length of the string, i.e. \[\mu = \dfrac{m}{l}\] and T is the applied tension.
Complete step by step solution:
If the length of the string is l , the mass of the string is m and the applied tension is T. Then the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
here \[\mu \] is the mass per unit length of the string, i.e. \[\Rightarrow \mu = \dfrac{m}{l}\]
If the density of the material of string is \[\rho \] then the mass per unit length of the string will be,
\[\mu = \dfrac{{\rho Al}}{l} = \rho A\]
So, the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{{\rho A}}} = \sqrt {\dfrac{T}{{\pi \rho {r^2}}}} \]
The wave speed on the string A will be,
\[{v_A} = \dfrac{1}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{\pi {\rho _A}}}} \]
The wave speed on the string B will be,
\[{v_B} = \dfrac{1}{{{r_B}}}\sqrt {\dfrac{{{T_B}}}{{\pi {\rho _B}}}} \]
It is given that the radius of A is double of the radius of B,
\[{r_A} = 2{r_B}\]
And both the strings are made of the same material,
\[{\rho _A} = {\rho _B}\]
The tension in both the strings are the same,
\[{T_A} = {T_B}\]
On dividing the wave speed of string A by the wave speed of string B, we get
\[\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{{T_B}}} \times \dfrac{{{\rho _B}}}{{{\rho _A}}}} \\ \]
\[\Rightarrow \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{2{r_B}}}\sqrt {1 \times 1} \\ \]
\[\therefore \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{1}{2}\]
Hence, the ratio of the speed of the transverse wave on the string A to the speed of the transverse wave on string B is equal to \[\dfrac{1}{2}\].
Therefore, the correct option is A.
Note: We can say from the obtained relation for the transverse wave and the density of the material that if we increase the density of the material the speed of the transverse wave decreases but the speed of longitudinal wave increases.
Formula used:
\[v = \sqrt {\dfrac{T}{\mu }} \]
where v is the speed of the transverse wave on a string with \[\mu \] as the mass per unit length of the string, i.e. \[\mu = \dfrac{m}{l}\] and T is the applied tension.
Complete step by step solution:
If the length of the string is l , the mass of the string is m and the applied tension is T. Then the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
here \[\mu \] is the mass per unit length of the string, i.e. \[\Rightarrow \mu = \dfrac{m}{l}\]
If the density of the material of string is \[\rho \] then the mass per unit length of the string will be,
\[\mu = \dfrac{{\rho Al}}{l} = \rho A\]
So, the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{{\rho A}}} = \sqrt {\dfrac{T}{{\pi \rho {r^2}}}} \]
The wave speed on the string A will be,
\[{v_A} = \dfrac{1}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{\pi {\rho _A}}}} \]
The wave speed on the string B will be,
\[{v_B} = \dfrac{1}{{{r_B}}}\sqrt {\dfrac{{{T_B}}}{{\pi {\rho _B}}}} \]
It is given that the radius of A is double of the radius of B,
\[{r_A} = 2{r_B}\]
And both the strings are made of the same material,
\[{\rho _A} = {\rho _B}\]
The tension in both the strings are the same,
\[{T_A} = {T_B}\]
On dividing the wave speed of string A by the wave speed of string B, we get
\[\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{{T_B}}} \times \dfrac{{{\rho _B}}}{{{\rho _A}}}} \\ \]
\[\Rightarrow \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{2{r_B}}}\sqrt {1 \times 1} \\ \]
\[\therefore \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{1}{2}\]
Hence, the ratio of the speed of the transverse wave on the string A to the speed of the transverse wave on string B is equal to \[\dfrac{1}{2}\].
Therefore, the correct option is A.
Note: We can say from the obtained relation for the transverse wave and the density of the material that if we increase the density of the material the speed of the transverse wave decreases but the speed of longitudinal wave increases.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

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 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
