
A wave of frequency \[{\rm{500Hz}}\] has a velocity of \[{\rm{360m}}{{\rm{s}}^{{\rm{ - 1}}}}\]. Calculate the distance between two points that are \[{60^0}\] out of phase.
A. 12 cm
B. 18 cm
C. 17 cm
D. 25 cm
Answer
146.7k+ views
Hint:To proceed with the problem, let’s see what the path difference and phase difference say. The distance travelled by the two waves from their respective sources to a given point on the pattern is known as path difference. The phase difference is defined as the difference in phase angle between two waves. Now we can solve the problem by considering the definitions.
Formula used:
The formula to find the wavelength is,
\[\lambda = \dfrac{v}{f}\]
Where, \[v\] is the velocity of the wave and \[f\] is the frequency of the wave.
To find the path difference, the formula is given by,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
Where, \[\Delta x\] is the path difference, \[\lambda \] is the wavelength and \[\Delta \varphi \] is the phase difference.
Complete step by step solution:
To find the path difference, first, we need to find the
\[\lambda = \dfrac{v}{f}\]
\[ \Rightarrow \lambda = \dfrac{{360}}{{500}}\]
By data we have \[{\rm{v = 360m}}{{\rm{s}}^{{\rm{ - 1}}}}\] and \[f = 500Hz\].
\[\lambda = 0.72\,m\]
Now let’s find the path difference or the distance between the two points from the above formula,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{{2\pi }} \times \dfrac{\pi }{3}\]
By data, \[\Delta \varphi = {60^0} = \dfrac{\pi }{3}\]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{6}\]
Now substitute the value of \[\lambda \]in the above equation we get,
\[\Delta x = \dfrac{{0.72}}{6}\]
\[ \Rightarrow \Delta x = 0.12\,m\]
Convert the above value from m to cm we get,
\[\therefore \Delta x = 12\,cm\]
Therefore, the distance between the two points is 12 cm.
Hence, option A is the correct answer.
Note:Now will see on what factors the frequency of a wave depends. As we know that the frequency is the number of cycles per second hence it depends only on the frequency of the source.
Formula used:
The formula to find the wavelength is,
\[\lambda = \dfrac{v}{f}\]
Where, \[v\] is the velocity of the wave and \[f\] is the frequency of the wave.
To find the path difference, the formula is given by,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
Where, \[\Delta x\] is the path difference, \[\lambda \] is the wavelength and \[\Delta \varphi \] is the phase difference.
Complete step by step solution:
To find the path difference, first, we need to find the
\[\lambda = \dfrac{v}{f}\]
\[ \Rightarrow \lambda = \dfrac{{360}}{{500}}\]
By data we have \[{\rm{v = 360m}}{{\rm{s}}^{{\rm{ - 1}}}}\] and \[f = 500Hz\].
\[\lambda = 0.72\,m\]
Now let’s find the path difference or the distance between the two points from the above formula,
\[\Delta x = \dfrac{\lambda }{{2\pi }}\Delta \varphi \]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{{2\pi }} \times \dfrac{\pi }{3}\]
By data, \[\Delta \varphi = {60^0} = \dfrac{\pi }{3}\]
\[ \Rightarrow \Delta x = \dfrac{\lambda }{6}\]
Now substitute the value of \[\lambda \]in the above equation we get,
\[\Delta x = \dfrac{{0.72}}{6}\]
\[ \Rightarrow \Delta x = 0.12\,m\]
Convert the above value from m to cm we get,
\[\therefore \Delta x = 12\,cm\]
Therefore, the distance between the two points is 12 cm.
Hence, option A is the correct answer.
Note:Now will see on what factors the frequency of a wave depends. As we know that the frequency is the number of cycles per second hence it depends only on the frequency of the source.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

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

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
