Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! - Oscillations and Waves - JEE Main-11

A particle executes simple harmonic motion and is located at x=a, b and c at
times t_{0},2t_{0},3t_{0}  respectively. The frequency of the oscillation is :

  • Option 1)

    \frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+c}{2b} \right )

  • Option 2)

    \frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+b}{2c} \right )

  • Option 3)

    \frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{2a+3c}{b} \right )

  • Option 4)

    \frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+2b}{3c} \right )

 

Answers (1)

x= A \sin (wt+\phi )

let \phi =0; x= A \sin wt

then a= A \sin wt_{0}

b= A \sin 2wt_{0}

c= A \sin 3wt_{0}

a+c = A (\sin wt_{0}+ \sin 3wt_{0})=2A \sin 2wt_{0} \cos (w t_{0})

a+c = b(2 \cos wt_{0})

\therefore w=\frac{t}{t_{0}} \cos ^{-1}(\frac{a+c}{2b})\Rightarrow f= \frac{1}{2\pi t_{0}} \cos ^{-1}\left ( \frac{a+c}{2b} \right )


Option 1)

\frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+c}{2b} \right )

This is correct

Option 2)

\frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+b}{2c} \right )

This is incorrect

Option 3)

\frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{2a+3c}{b} \right )

This is incorrect

Option 4)

\frac{1}{2\pi t_{0}}\cos ^{-1}\left ( \frac{a+2b}{3c} \right )

This is incorrect

Posted by

subam

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in

Latest Question