Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line from the same mean positon. The maximum distance between the two particles is a <img alt="\s

Two particles P and Q describe S.H.M. of same amplitude a, same frequency f along the same straight line from the same mean positon. The maximum distance between the two particles is a \sqrt{2}. The phase difference between the particle is:      

Option: 1

Zero


Option: 2

\frac{\pi }{2}


Option: 3

\frac{\pi }{6}


Option: 4

\frac{\pi }{3}


Answers (1)

best_answer

 

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

x_{1}= a\sin \left ( \omega t +\phi_{1} \right )

x_{2}= a\sin \left ( \omega t +\phi_{2} \right )\Rightarrow \left | x_{1}-x_{2} \right | = 2a\sin \left ( \omega t +\frac{\phi _{1}+\phi _{2}}{2} \right )\cos \left ( \frac{\phi _{1}-\phi_{2}}{2} \right )

To maximize \left | x_{1}-x_{2} \right | :

\sin \left ( \omega t +\frac{\phi _{1}+\phi _{2}}{2} \right )=1

\Rightarrow a\sqrt{2}=2a \times 1\cos \left ( \frac{\phi _{1}-\phi _{2}}{2} \right )

\Rightarrow \frac{1}{\sqrt{2}} = \cos \left ( \frac{\phi _{1}-\phi_{2}}{2} \right ) \Rightarrow \frac{\pi}{4}=\frac{\phi _{1}-\phi_{2}}{2}

\phi_{1}-\phi_{2}=\frac{\pi}{2}

Posted by

Rakesh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel
anam-khan

CSAB 2025 counselling dates out for special, supernumerary rounds through JEE Mains

Latest Question