The distance between three consecutive crests of waves is $60cm$. If the waves are produced at the rate of $180/\min $, calculate (i) wavelength (ii) time period (iii) wave velocity.
Answer
265.2k+ views
Hint: Using the definition of the wavelength, we can determine its value from the given distance. Also, from the given rate of wave production we can determine the time period. Finally, using the formula for the wave velocity in terms of wavelength and the time period we can determine its value.
Formula used: The formula used to solve this question is given by
$v = \dfrac{\lambda }{T}$, here $v$ is the velocity of a wave having a wavelength of $\lambda $ and a time period of $T$.
Complete step-by-step solution:
(i) We know that the distance between two consecutive crests or two consecutive troughs of a wave is equal to the wavelength of the wave. So the distance between the three consecutive crests of the wave must be equal to the twice of the wavelength. According to the question, we are given the distance between three consecutive to be $60cm$. So if the wavelength of the given wave is equal to $\lambda $, then we can say that
$2\lambda = 60cm$
$ \Rightarrow \lambda = 30cm$
We know that $1cm = 0.01m$. So we get
$\lambda = 0.3m$..............(1)
Hence, the wavelength of the given wave is equal to $0.3m$.
(ii) According to the question, the waves are produced at the rate of $180/\min $. This means that $180$ waves are produced in a minute. Since there are $60s$ in a minute, so the number of waves produced in one second becomes
$n = \dfrac{{180}}{{60}}{s^{ - 1}}$
$ \Rightarrow n = 3{s^{ - 1}}$
Now, we know that the frequency of a wave is equal to the number of cycles produced in a second. Since one cycle corresponds to one wave, we can say that the frequency is equal to the number of waves produced in one second. So the above number is nothing but the frequency of the given wave. So we have
$f = 3{s^{ - 1}}$
We know that the time period is equal to the inverse of the frequency, that is,
$T = \dfrac{1}{f}$
$ \Rightarrow T = \dfrac{1}{3}s$ …………………………..(2)
Hence, the time period of the given wave is equal to $\dfrac{1}{3}s$.
(iii) We know that the velocity of a wave is equal to the ratio of the wavelength and the time period, that is,
$v = \dfrac{\lambda }{T}$
Substituting (1) and (2) in the above equation, we get
$v = \dfrac{{0.3}}{{1/3}}$
$ \Rightarrow v = 0.9m{s^{ - 1}}$
Hence, the velocity of the given wave is equal to $0.9m{s^{ - 1}}$.
Note: Do not take the distance between three consecutive crests to be equal to thrice the wavelength of the wave. Remember, two consecutive crests or troughs correspond to one wave. So three consecutive crests corresponds to two waves, and hence twice the wavelength.
Formula used: The formula used to solve this question is given by
$v = \dfrac{\lambda }{T}$, here $v$ is the velocity of a wave having a wavelength of $\lambda $ and a time period of $T$.
Complete step-by-step solution:
(i) We know that the distance between two consecutive crests or two consecutive troughs of a wave is equal to the wavelength of the wave. So the distance between the three consecutive crests of the wave must be equal to the twice of the wavelength. According to the question, we are given the distance between three consecutive to be $60cm$. So if the wavelength of the given wave is equal to $\lambda $, then we can say that
$2\lambda = 60cm$
$ \Rightarrow \lambda = 30cm$
We know that $1cm = 0.01m$. So we get
$\lambda = 0.3m$..............(1)
Hence, the wavelength of the given wave is equal to $0.3m$.
(ii) According to the question, the waves are produced at the rate of $180/\min $. This means that $180$ waves are produced in a minute. Since there are $60s$ in a minute, so the number of waves produced in one second becomes
$n = \dfrac{{180}}{{60}}{s^{ - 1}}$
$ \Rightarrow n = 3{s^{ - 1}}$
Now, we know that the frequency of a wave is equal to the number of cycles produced in a second. Since one cycle corresponds to one wave, we can say that the frequency is equal to the number of waves produced in one second. So the above number is nothing but the frequency of the given wave. So we have
$f = 3{s^{ - 1}}$
We know that the time period is equal to the inverse of the frequency, that is,
$T = \dfrac{1}{f}$
$ \Rightarrow T = \dfrac{1}{3}s$ …………………………..(2)
Hence, the time period of the given wave is equal to $\dfrac{1}{3}s$.
(iii) We know that the velocity of a wave is equal to the ratio of the wavelength and the time period, that is,
$v = \dfrac{\lambda }{T}$
Substituting (1) and (2) in the above equation, we get
$v = \dfrac{{0.3}}{{1/3}}$
$ \Rightarrow v = 0.9m{s^{ - 1}}$
Hence, the velocity of the given wave is equal to $0.9m{s^{ - 1}}$.
Note: Do not take the distance between three consecutive crests to be equal to thrice the wavelength of the wave. Remember, two consecutive crests or troughs correspond to one wave. So three consecutive crests corresponds to two waves, and hence twice the wavelength.
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

Understanding Atomic Structure for Beginners

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

JEE Advanced Marks vs Rank 2025 - Predict Your IIT Rank Based on Score

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

