The equation of a travelling sound wave is $y = 6.0\sin (600t - 1.8x)$ where $y$ is measured in ${10^{ - 5\,}}\,m$, $t$ in second and $x$ in meter.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particles to the wave speed.
Answer
264.9k+ views
Hint: The equation for the oscillating wave is given as $y = A\sin (\omega t - K) + b$. where. $A$ = Displacement amplitude, $\omega $ is the wavelength, $t$ is the time, $K$ is some arbitrary constant representing time offset, $b$ is arbitrary displacement offset. Comparing the given equation with this equation, we can easily find the given values. Ratio of displacement is given as $\dfrac{A}{\lambda }$. The maximum velocity of a particle is known as velocity amplitude. The wave speed is calculated using the relation between velocity, frequency and wavelength.
Complete step by step solution:
The given relation is: $y = 6.0\sin (600t - 1.8x)$ comparing this with the equation of oscillating wave, $x = A\sin (\omega t - K) + b$ we have:
\[A = 6.0\]
\[\omega = 600\]
\[K = 1.8\]
Since, we are given that the displacement is in the order of ${10^{ - 5\,}}\,m$. Therefore, we can have the amplitude as:
\[A = 6.0 \times {10^{ - 5}}\,m\]
The wavelength \[\lambda \] is given as:
\[\lambda = \dfrac{{2\pi }}{K}\]
Substituting the values, we get:
\[\lambda = \dfrac{{2\pi }}{{1.8}}\] …………………………………...equation \[(1)\]
The displacement amplitude is given as \[\dfrac{A}{\lambda }\] , substituting the value of amplitude and wavelength, we get:
\[\dfrac{A}{\lambda } = \dfrac{{6 \times {{10}^{ - 5}}}}{{\dfrac{{2\pi }}{{1.8}}}}\]
\[ \Rightarrow \dfrac{A}{\lambda } = \dfrac{{5.4 \times {{10}^{ - 5}}}}{\pi }\]
\[ \Rightarrow \dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}\]
Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is \[1.7 \times {10^{ - 5}}\]
To find the ratio of the velocity amplitude of the particles to the wave speed, let us find the velocity of the particle and the wave speed.
As, velocity is rate of change of displacement thus, differentiating the equation of displacement we will get the velocity of the particle.
\[v = \dfrac{{dy}}{{dt}}\]
Here, \[v\] is the velocity of the particle.
\[ \Rightarrow v = \dfrac{{d\left( {6.0\sin (600t - 1.8x)} \right)}}{{dt}}\]
\[ \Rightarrow v = 3600\cos (600t - 1.8x) \times {10^{ - 5}}\]
This velocity will be maximum when the value of cosine is maximum. The maximum value of cosine is $1$ . Therefore, the maximum velocity will be
\[ \Rightarrow v = 3600(1) \times {10^{ - 5}}\]
\[ \Rightarrow v = 3600 \times {10^{ - 5}}\,m\,{s^{ - 1}}\] ………………………….equation \[(2)\]
Now, for the speed of the wave, we have
\[\omega = 600\]
But \[\omega = 2\pi f\] , \[f\] is the frequency. Thus, the frequency will be:
\[f = \dfrac{\omega }{{2\pi }}\]
Substituting the values, we get
\[f = \dfrac{{600}}{{2\pi }}\]
The wave speed \[{v_s}\] is given \[{v_s} = f \times \lambda \] .
\[ \Rightarrow {v_s} = \dfrac{{600}}{{2\pi }} \times \dfrac{{2\pi }}{{1.8}}\]
\[ \Rightarrow {v_s} = \dfrac{{1000}}{3}\,m\,{s^{ - 1}}\] …………………..equation \[(3)\]
Dividing equation \[2\] by equation \[3\] , we will get the ratio of the velocity amplitude of the particles to the wave speed.
\[\dfrac{v}{{{v_s}}} = \dfrac{{3600 \times {{10}^{ - 5}}}}{{\dfrac{{1000}}{3}}}\]
\[\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}\]
Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is \[\dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}\] and ratio of the velocity amplitude of the particles to the wave speed is \[\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}\].
Note: The ratio will be a dimensionless quantity. Comparing the term with the general equation, we get the value of various variables. The equation of the velocity of the particle is obtained by differentiating the equation of the displacement with respect to time. The wave speed is calculated using the relation between speed, frequency and wavelength. The magnitude of displacement is given as ${10^{ - 5\,}}\,m$.
Complete step by step solution:
The given relation is: $y = 6.0\sin (600t - 1.8x)$ comparing this with the equation of oscillating wave, $x = A\sin (\omega t - K) + b$ we have:
\[A = 6.0\]
\[\omega = 600\]
\[K = 1.8\]
Since, we are given that the displacement is in the order of ${10^{ - 5\,}}\,m$. Therefore, we can have the amplitude as:
\[A = 6.0 \times {10^{ - 5}}\,m\]
The wavelength \[\lambda \] is given as:
\[\lambda = \dfrac{{2\pi }}{K}\]
Substituting the values, we get:
\[\lambda = \dfrac{{2\pi }}{{1.8}}\] …………………………………...equation \[(1)\]
The displacement amplitude is given as \[\dfrac{A}{\lambda }\] , substituting the value of amplitude and wavelength, we get:
\[\dfrac{A}{\lambda } = \dfrac{{6 \times {{10}^{ - 5}}}}{{\dfrac{{2\pi }}{{1.8}}}}\]
\[ \Rightarrow \dfrac{A}{\lambda } = \dfrac{{5.4 \times {{10}^{ - 5}}}}{\pi }\]
\[ \Rightarrow \dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}\]
Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is \[1.7 \times {10^{ - 5}}\]
To find the ratio of the velocity amplitude of the particles to the wave speed, let us find the velocity of the particle and the wave speed.
As, velocity is rate of change of displacement thus, differentiating the equation of displacement we will get the velocity of the particle.
\[v = \dfrac{{dy}}{{dt}}\]
Here, \[v\] is the velocity of the particle.
\[ \Rightarrow v = \dfrac{{d\left( {6.0\sin (600t - 1.8x)} \right)}}{{dt}}\]
\[ \Rightarrow v = 3600\cos (600t - 1.8x) \times {10^{ - 5}}\]
This velocity will be maximum when the value of cosine is maximum. The maximum value of cosine is $1$ . Therefore, the maximum velocity will be
\[ \Rightarrow v = 3600(1) \times {10^{ - 5}}\]
\[ \Rightarrow v = 3600 \times {10^{ - 5}}\,m\,{s^{ - 1}}\] ………………………….equation \[(2)\]
Now, for the speed of the wave, we have
\[\omega = 600\]
But \[\omega = 2\pi f\] , \[f\] is the frequency. Thus, the frequency will be:
\[f = \dfrac{\omega }{{2\pi }}\]
Substituting the values, we get
\[f = \dfrac{{600}}{{2\pi }}\]
The wave speed \[{v_s}\] is given \[{v_s} = f \times \lambda \] .
\[ \Rightarrow {v_s} = \dfrac{{600}}{{2\pi }} \times \dfrac{{2\pi }}{{1.8}}\]
\[ \Rightarrow {v_s} = \dfrac{{1000}}{3}\,m\,{s^{ - 1}}\] …………………..equation \[(3)\]
Dividing equation \[2\] by equation \[3\] , we will get the ratio of the velocity amplitude of the particles to the wave speed.
\[\dfrac{v}{{{v_s}}} = \dfrac{{3600 \times {{10}^{ - 5}}}}{{\dfrac{{1000}}{3}}}\]
\[\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}\]
Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is \[\dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}\] and ratio of the velocity amplitude of the particles to the wave speed is \[\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}\].
Note: The ratio will be a dimensionless quantity. Comparing the term with the general equation, we get the value of various variables. The equation of the velocity of the particle is obtained by differentiating the equation of the displacement with respect to time. The wave speed is calculated using the relation between speed, frequency and wavelength. The magnitude of displacement is given as ${10^{ - 5\,}}\,m$.
Recently Updated Pages
JEE Main Mock Test 2025-26: Principles Related To Practical

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main Statistics and Probability Mock Test 2025-26

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

